x
4
14
+
x
3
14
−
13
x
2
14
−
x
14
+
19
14
{\displaystyle {\tfrac {{x}^{4}}{14}}+{\tfrac {{x}^{3}}{14}}-{\tfrac {13{{x}^{2}}}{14}}-{\tfrac {x}{14}}+{\tfrac {19}{14}}}
x tengellyel való metszéspontok a függvény zérushelyei (y = 0).A negyedfokú egyenlet olyan egyenlet melynek az egyik oldalán lévő kifejezés egy negyedfokú polinomfüggvény , a másik oldalán lévő kifejezés pedig zéró. Általános alakja:
a
⋅
x
4
+
b
⋅
x
3
+
c
⋅
x
2
+
d
⋅
x
+
e
=
0
{\displaystyle a\cdot x^{4}+b\cdot x^{3}+c\cdot x^{2}+d\cdot x+e=0\,}
Megoldását Gerolamo Cardano inasa és tanítványa, Lodovico Ferrari (1522-1565) fedezte fel; a megoldás Cardano Ars magna című munkájában jelent meg.
Ez a legmagasabb fokú egyenlet, amely általános alakban megoldható; ezt Niels Henrik Abel bizonyította be 1824-ben.
Az általános negyedfokú egyenlet gyökei Ha
Δ
≥
0
{\displaystyle \Delta \geq 0}
(
B
=
0
)
{\displaystyle \left(B=0\right)}
(
A
>
0
)
{\displaystyle \left(A>0\right)}
(
C
=
A
2
4
)
{\displaystyle \left(C={\frac {{A}^{2}}{4}}\right)}
x
1
,
2
=
−
b
4
a
+
i
⋅
A
2
x
3
,
4
=
−
b
4
a
−
i
⋅
A
2
{\displaystyle {\begin{aligned}&{{x}_{1,2}}=-{\frac {b}{4a}}+i\cdot {\sqrt {\frac {A}{2}}}\\&{{x}_{3,4}}=-{\frac {b}{4a}}-i\cdot {\sqrt {\frac {A}{2}}}\\\end{aligned}}}
ellenkező esetben:
x
1
,
2
=
−
b
4
a
+
s
i
g
(
−
B
)
−
A
6
+
u
+
v
±
−
A
3
−
(
u
+
v
)
+
(
A
6
+
2
(
u
+
v
)
)
2
−
C
x
3
,
4
=
−
b
4
a
−
s
i
g
(
−
B
)
−
A
6
+
u
+
v
±
i
⋅
A
3
+
(
u
+
v
)
+
(
A
6
+
2
(
u
+
v
)
)
2
−
C
{\displaystyle {\begin{aligned}&{{x}_{1,2}}=-{\frac {b}{4a}}+sig\left(-B\right){\sqrt {-{\frac {A}{6}}+u+v}}\pm {\sqrt {-{\frac {A}{3}}-\left(u+v\right)+{\sqrt {{{\left({\frac {A}{6}}+2\left(u+v\right)\right)}^{2}}-C}}}}\\&{{x}_{3,4}}=-{\frac {b}{4a}}-sig\left(-B\right){\sqrt {-{\frac {A}{6}}+u+v}}\pm i\cdot {\sqrt {{\frac {A}{3}}+\left(u+v\right)+{\sqrt {{{\left({\frac {A}{6}}+2\left(u+v\right)\right)}^{2}}-C}}}}\\\end{aligned}}}
Ha
Δ
<
0
{\displaystyle \Delta <0}
(
C
>
A
2
4
)
{\displaystyle \left(C>{\frac {{A}^{2}}{4}}\right)}
(
A
>
0
)
{\displaystyle \left(A>0\right)}
x
1
,
2
=
−
b
4
a
−
s
i
g
(
−
B
)
Y
1
±
i
⋅
(
−
Y
2
+
−
Y
3
)
x
3
,
4
=
−
b
4
a
+
s
i
g
(
−
B
)
Y
1
±
i
⋅
(
−
Y
2
−
−
Y
3
)
{\displaystyle {\begin{aligned}&{{x}_{1,2}}=-{\frac {b}{4a}}-sig\left(-B\right){\sqrt {{Y}_{1}}}\pm i\cdot \left({\sqrt {-{{Y}_{2}}}}+{\sqrt {-{{Y}_{3}}}}\right)\\&{{x}_{3,4}}=-{\frac {b}{4a}}+sig\left(-B\right){\sqrt {{Y}_{1}}}\pm i\cdot \left({\sqrt {-{{Y}_{2}}}}-{\sqrt {-{{Y}_{3}}}}\right)\\\end{aligned}}}
ellenkező esetben mind a négy gyök valós:
x
1
,
2
=
−
b
4
a
+
s
i
g
(
−
B
)
Y
1
±
(
Y
2
+
Y
3
)
x
3
,
4
=
−
b
4
a
−
s
i
g
(
−
B
)
Y
1
±
(
Y
2
−
Y
3
)
{\displaystyle {\begin{aligned}&{{x}_{1,2}}=-{\frac {b}{4a}}+sig\left(-B\right){\sqrt {{Y}_{1}}}\pm \left({\sqrt {{Y}_{2}}}+{\sqrt {{Y}_{3}}}\right)\\&{{x}_{3,4}}=-{\frac {b}{4a}}-sig\left(-B\right){\sqrt {{Y}_{1}}}\pm \left({\sqrt {{Y}_{2}}}-{\sqrt {{Y}_{3}}}\right)\\\end{aligned}}}
A
=
−
3
b
2
8
a
2
+
c
a
{\displaystyle A=-{\frac {3{{b}^{2}}}{8{{a}^{2}}}}+{\frac {c}{a}}}
B
=
b
3
8
a
3
−
b
c
2
a
2
+
d
a
{\displaystyle B={\frac {{b}^{3}}{8{{a}^{3}}}}-{\frac {bc}{2{{a}^{2}}}}+{\frac {d}{a}}}
C
=
−
3
b
4
256
a
4
+
b
2
c
16
a
3
−
b
d
4
a
2
+
e
a
{\displaystyle C=-{\frac {3{{b}^{4}}}{256{{a}^{4}}}}+{\frac {{{b}^{2}}c}{16{{a}^{3}}}}-{\frac {bd}{4{{a}^{2}}}}+{\frac {e}{a}}}
P
=
−
A
2
48
−
C
4
{\displaystyle P=-{\frac {{A}^{2}}{48}}-{\frac {C}{4}}}
Q
=
−
A
3
864
−
B
2
64
+
A
C
24
{\displaystyle Q=-{\frac {{A}^{3}}{864}}-{\frac {{B}^{2}}{64}}+{\frac {AC}{24}}}
Δ
=
(
Q
2
)
2
+
(
P
3
)
3
{\displaystyle \Delta ={{\left({\frac {Q}{2}}\right)}^{2}}+{{\left({\frac {P}{3}}\right)}^{3}}}
u
,
v
=
−
Q
2
±
Δ
3
{\displaystyle u,v={\sqrt[{3}]{-{\frac {Q}{2}}\pm {\sqrt {\Delta }}}}}
Y
k
=
−
A
6
+
2
−
P
/
3
⋅
cos
(
2
(
k
−
1
)
⋅
π
3
+
1
3
⋅
arccos
−
Q
/
2
−
(
P
/
3
)
3
)
{\displaystyle {{Y}_{k}}=-{\frac {A}{6}}+2{\sqrt {-P/3}}\cdot \cos \left({\frac {2\left(k-1\right)\cdot \pi }{3}}+{\frac {1}{3}}\cdot \arccos {\frac {-Q/2}{\sqrt {-{{\left(P/3\right)}^{3}}}}}\right)}
s
i
g
(
x
)
=
{
+
1
,
x
≥
0
−
1
,
x
<
0
{\displaystyle sig\left(x\right)=\left\{{\begin{aligned}&+1,x\geq 0\\&-1,x<0\\\end{aligned}}\right.}
Viète-formulák
x
1
+
x
2
+
x
3
+
x
4
=
−
b
a
{\displaystyle x_{1}+x_{2}+x_{3}+x_{4}=-{\frac {b}{a}}}
x
1
x
2
+
x
1
x
3
+
x
1
x
4
+
x
2
x
3
+
x
2
x
4
+
x
3
x
4
=
c
a
{\displaystyle x_{1}x_{2}+x_{1}x_{3}+x_{1}x_{4}+x_{2}x_{3}+x_{2}x_{4}+x_{3}x_{4}={\frac {c}{a}}}
x
1
x
2
x
3
+
x
1
x
2
x
4
+
x
1
x
3
x
4
+
x
2
x
3
x
4
=
−
d
a
{\displaystyle x_{1}x_{2}x_{3}+x_{1}x_{2}x_{4}+x_{1}x_{3}x_{4}+x_{2}x_{3}x_{4}=-{\frac {d}{a}}}
x
1
x
2
x
3
x
4
=
e
a
{\displaystyle x_{1}x_{2}x_{3}x_{4}={\frac {e}{a}}}
Az általános negyedfokú egyenlet megoldása Mivel
(
y
1
+
y
2
+
y
3
)
4
−
2
(
y
1
2
+
y
2
2
+
y
3
2
)
(
y
1
+
y
2
+
y
3
)
2
−
8
y
1
y
2
y
3
(
y
1
+
y
2
+
y
3
)
+
(
y
1
2
+
y
2
2
+
y
3
2
)
2
−
4
(
y
1
2
y
2
2
+
y
1
2
y
3
2
+
y
2
2
y
3
2
)
=
0
{\displaystyle {{\left({{y}_{1}}+{{y}_{2}}+{{y}_{3}}\right)}^{4}}-2\left(y_{1}^{2}+y_{2}^{2}+y_{3}^{2}\right){{\left({{y}_{1}}+{{y}_{2}}+{{y}_{3}}\right)}^{2}}-8{{y}_{1}}{{y}_{2}}{{y}_{3}}\left({{y}_{1}}+{{y}_{2}}+{{y}_{3}}\right)+{{\left(y_{1}^{2}+y_{2}^{2}+y_{3}^{2}\right)}^{2}}-4\left(y_{1}^{2}y_{2}^{2}+y_{1}^{2}y_{3}^{2}+y_{2}^{2}y_{3}^{2}\right)=0}
X
4
−
2
(
y
1
2
+
y
2
2
+
y
3
2
)
⋅
X
2
−
8
y
1
y
2
y
3
⋅
X
+
(
y
1
2
+
y
2
2
+
y
3
2
)
2
−
4
(
y
1
2
y
2
2
+
y
1
2
y
3
2
+
y
2
2
y
3
2
)
=
0
{\displaystyle {{X}^{4}}-2\left(y_{1}^{2}+y_{2}^{2}+y_{3}^{2}\right)\cdot {{X}^{2}}-8{{y}_{1}}{{y}_{2}}{{y}_{3}}\cdot X+{{\left(y_{1}^{2}+y_{2}^{2}+y_{3}^{2}\right)}^{2}}-4\left(y_{1}^{2}y_{2}^{2}+y_{1}^{2}y_{3}^{2}+y_{2}^{2}y_{3}^{2}\right)=0}
X
1
=
y
1
+
y
2
+
y
3
{\displaystyle {{X}_{1}}={{y}_{1}}+{{y}_{2}}+{{y}_{3}}}
Ez igaz marad akkor is ha
X
=
y
1
±
(
y
2
+
y
3
)
{\displaystyle X={{y}_{1}}\pm \left({{y}_{2}}+{{y}_{3}}\right)}
X
=
−
y
1
±
(
y
2
−
y
3
)
{\displaystyle X=-{{y}_{1}}\pm \left({{y}_{2}}-{{y}_{3}}\right)}
X
4
−
2
(
y
1
2
+
y
2
2
+
y
3
2
)
⋅
X
2
−
8
y
1
y
2
y
3
⋅
X
+
(
y
1
2
+
y
2
2
+
y
3
2
)
2
−
4
(
y
1
2
y
2
2
+
y
1
2
y
3
2
+
y
2
2
y
3
2
)
=
0
{\displaystyle {{X}^{4}}-2\left(y_{1}^{2}+y_{2}^{2}+y_{3}^{2}\right)\cdot {{X}^{2}}-8{{y}_{1}}{{y}_{2}}{{y}_{3}}\cdot X+{{\left(y_{1}^{2}+y_{2}^{2}+y_{3}^{2}\right)}^{2}}-4\left(y_{1}^{2}y_{2}^{2}+y_{1}^{2}y_{3}^{2}+y_{2}^{2}y_{3}^{2}\right)=0}
X
1
,
2
=
+
y
1
±
(
y
2
+
y
3
)
X
3
,
4
=
−
y
1
±
(
y
2
−
y
3
)
{\displaystyle {\begin{aligned}&{{X}_{1,2}}=+{{y}_{1}}\pm \left({{y}_{2}}+{{y}_{3}}\right)\\&{{X}_{3,4}}=-{{y}_{1}}\pm \left({{y}_{2}}-{{y}_{3}}\right)\\\end{aligned}}}
X
4
+
A
⋅
X
2
+
B
⋅
X
+
C
=
0
{\displaystyle {{X}^{4}}+A\cdot {{X}^{2}}+B\cdot X+C=0}
{
−
2
(
y
1
2
+
y
2
2
+
y
3
2
)
=
A
−
8
y
1
y
2
y
3
=
B
(
y
1
2
+
y
2
2
+
y
3
2
)
2
−
4
(
y
1
2
y
2
2
+
y
1
2
y
3
2
+
y
2
2
y
3
2
)
=
C
{\displaystyle \left\{{\begin{aligned}&-2\left(y_{1}^{2}+y_{2}^{2}+y_{3}^{2}\right)=A\\&-8{{y}_{1}}{{y}_{2}}{{y}_{3}}=B\\&{{\left(y_{1}^{2}+y_{2}^{2}+y_{3}^{2}\right)}^{2}}-4\left(y_{1}^{2}y_{2}^{2}+y_{1}^{2}y_{3}^{2}+y_{2}^{2}y_{3}^{2}\right)=C\\\end{aligned}}\right.}
a
,
b
,
c
{\displaystyle a,b,c}
A
,
B
,
C
{\displaystyle A,B,C}
{
y
1
2
+
y
2
2
+
y
3
2
=
−
A
2
y
1
2
y
2
2
+
y
1
2
y
3
2
+
y
2
2
y
3
2
=
A
2
16
−
C
4
y
1
2
y
2
2
y
3
2
=
B
2
64
{\displaystyle \left\{{\begin{aligned}&y_{1}^{2}+y_{2}^{2}+y_{3}^{2}=-{\frac {A}{2}}\\&y_{1}^{2}y_{2}^{2}+y_{1}^{2}y_{3}^{2}+y_{2}^{2}y_{3}^{2}={\frac {{A}^{2}}{16}}-{\frac {C}{4}}\\&y_{1}^{2}y_{2}^{2}y_{3}^{2}={\frac {{B}^{2}}{64}}\\\end{aligned}}\right.}
(
y
2
)
3
+
A
2
⋅
(
y
2
)
2
+
(
A
2
16
−
C
4
)
⋅
(
y
2
)
−
(
B
8
)
2
=
0
{\displaystyle {{\left({{y}^{2}}\right)}^{3}}+{\frac {A}{2}}\cdot {{\left({{y}^{2}}\right)}^{2}}+\left({\frac {{A}^{2}}{16}}-{\frac {C}{4}}\right)\cdot \left({{y}^{2}}\right)-{{\left({\frac {B}{8}}\right)}^{2}}=0}
y
1
y
2
y
3
=
−
B
8
{\displaystyle {{y}_{1}}{{y}_{2}}{{y}_{3}}=-{\frac {B}{8}}}
P
=
−
A
2
48
−
C
4
Q
=
−
A
3
864
−
B
2
64
+
A
C
24
{\displaystyle {\begin{aligned}&P=-{\frac {{A}^{2}}{48}}-{\frac {C}{4}}\\&Q=-{\frac {{A}^{3}}{864}}-{\frac {{B}^{2}}{64}}+{\frac {AC}{24}}\\\end{aligned}}}
Δ
=
(
Q
2
)
2
+
(
P
3
)
3
u
,
v
=
−
Q
2
±
Δ
3
{\displaystyle {\begin{aligned}&\Delta ={{\left({\frac {Q}{2}}\right)}^{2}}+{{\left({\frac {P}{3}}\right)}^{3}}\\&u,v={\sqrt[{3}]{-{\frac {Q}{2}}\pm {\sqrt {\Delta }}}}\\\end{aligned}}}
Δ
≥
0
{\displaystyle \Delta \geq 0}
y
1
2
=
−
A
6
+
u
+
v
y
2
,
3
2
=
−
A
6
−
u
+
v
2
±
i
(
u
−
v
)
3
2
{\displaystyle {\begin{aligned}&y_{1}^{2}=-{\frac {A}{6}}+u+v\\&y_{2,3}^{2}=-{\frac {A}{6}}-{\frac {u+v}{2}}\pm i{\frac {\left(u-v\right){\sqrt {3}}}{2}}\\\end{aligned}}}
y
1
=
−
A
6
+
u
+
v
{\displaystyle {{y}_{1}}={\sqrt {-{\frac {A}{6}}+u+v}}}
y
2
,
3
{\displaystyle {{y}_{2,3}}}
α
±
i
⋅
β
=
±
(
α
+
α
2
+
β
2
2
±
i
⋅
s
i
g
(
b
)
⋅
−
α
+
α
2
+
β
2
2
)
{\displaystyle {\sqrt {\alpha \pm i\cdot \beta }}=\pm \left({\sqrt {\frac {\alpha +{\sqrt {{{\alpha }^{2}}+{{\beta }^{2}}}}}{2}}}\pm i\cdot sig\left(b\right)\cdot {\sqrt {\frac {-\alpha +{\sqrt {{{\alpha }^{2}}+{{\beta }^{2}}}}}{2}}}\right)}
ennek eredményeként:
y
2
,
3
=
1
2
−
A
3
−
(
u
+
v
)
+
(
A
6
+
2
(
u
+
v
)
)
2
−
C
±
i
2
⋅
A
3
+
(
u
+
v
)
+
(
A
6
+
2
(
u
+
v
)
)
2
−
C
{\displaystyle {{y}_{2,3}}={\frac {1}{2}}{\sqrt {-{\frac {A}{3}}-\left(u+v\right)+{\sqrt {{{\left({\frac {A}{6}}+2\left(u+v\right)\right)}^{2}}-C}}}}\pm {\frac {i}{2}}\cdot {\sqrt {{\frac {A}{3}}+\left(u+v\right)+{\sqrt {{{\left({\frac {A}{6}}+2\left(u+v\right)\right)}^{2}}-C}}}}}
y
2
⋅
y
3
=
1
2
(
A
6
+
2
(
u
+
v
)
)
2
−
C
≥
0
{\displaystyle {{y}_{2}}\cdot {{y}_{3}}={\frac {1}{2}}{\sqrt {{{\left({\frac {A}{6}}+2\left(u+v\right)\right)}^{2}}-C}}\geq 0}
y
1
y
2
y
3
=
−
B
8
{\displaystyle {{y}_{1}}{{y}_{2}}{{y}_{3}}=-{\frac {B}{8}}}
y
1
=
s
i
g
(
−
B
)
−
A
6
+
u
+
v
{\displaystyle {{y}_{1}}=sig\left(-B\right){\sqrt {-{\frac {A}{6}}+u+v}}}
Tehát pozitív delta esetén a gyökok:
X
1
,
2
=
+
s
i
g
(
−
B
)
−
A
6
+
u
+
v
±
−
A
3
−
(
u
+
v
)
+
(
A
6
+
2
(
u
+
v
)
)
2
−
C
X
3
,
4
=
−
s
i
g
(
−
B
)
−
A
6
+
u
+
v
±
i
⋅
A
3
+
(
u
+
v
)
+
(
A
6
+
2
(
u
+
v
)
)
2
−
C
{\displaystyle {\begin{aligned}&{{X}_{1,2}}=+sig\left(-B\right){\sqrt {-{\frac {A}{6}}+u+v}}\pm {\sqrt {-{\frac {A}{3}}-\left(u+v\right)+{\sqrt {{{\left({\frac {A}{6}}+2\left(u+v\right)\right)}^{2}}-C}}}}\\&{{X}_{3,4}}=-sig\left(-B\right){\sqrt {-{\frac {A}{6}}+u+v}}\pm i\cdot {\sqrt {{\frac {A}{3}}+\left(u+v\right)+{\sqrt {{{\left({\frac {A}{6}}+2\left(u+v\right)\right)}^{2}}-C}}}}\\\end{aligned}}}
B
=
0
{\displaystyle B=0}
A
>
0
{\displaystyle A>0}
C
=
A
2
4
{\displaystyle C={\frac {{A}^{2}}{4}}}
−
A
6
+
u
+
v
<
0
{\displaystyle -{\frac {A}{6}}+u+v<0}
y
1
{\displaystyle {{y}_{1}}}
X
1
,
2
=
+
i
⋅
A
2
X
3
,
4
=
−
i
⋅
A
2
{\displaystyle {\begin{aligned}&{{X}_{1,2}}=+i\cdot {\sqrt {\frac {A}{2}}}\\&{{X}_{3,4}}=-i\cdot {\sqrt {\frac {A}{2}}}\\\end{aligned}}}
Δ
<
0
{\displaystyle \Delta <0}
y
k
=
±
−
A
6
+
2
−
P
/
3
⋅
cos
(
2
(
k
−
1
)
⋅
π
3
+
1
3
⋅
arccos
−
Q
/
2
−
(
P
/
3
)
3
)
{\displaystyle {{y}_{k}}=\pm {\sqrt {-{\frac {A}{6}}+2{\sqrt {-P/3}}\cdot \cos \left({\frac {2\left(k-1\right)\cdot \pi }{3}}+{\frac {1}{3}}\cdot \arccos {\frac {-Q/2}{\sqrt {-{{\left(P/3\right)}^{3}}}}}\right)}}}
(
4
⋅
C
>
A
2
)
{\displaystyle \left(4\cdot C>{{A}^{2}}\right)}
(
A
>
0
)
{\displaystyle \left(A>0\right)}
y
2
,
3
{\displaystyle {{y}_{2,3}}}
y
1
y
2
y
3
=
−
B
8
{\displaystyle {{y}_{1}}{{y}_{2}}{{y}_{3}}=-{\frac {B}{8}}}
s
i
g
(
−
B
)
{\displaystyle sig\left(-B\right)}
X
1
,
2
=
−
s
i
g
(
−
B
)
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{\displaystyle {\begin{aligned}&{{X}_{1,2}}=-sig\left(-B\right)\cdot {{y}_{1}}\pm i\cdot \left({\sqrt {-y_{2}^{2}}}+{\sqrt {-y_{3}^{2}}}\right)\\&{{X}_{3,4}}=+sig\left(-B\right)\cdot {{y}_{1}}\pm i\cdot \left({\sqrt {-y_{2}^{2}}}-{\sqrt {-y_{3}^{2}}}\right)\\\end{aligned}}}
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{\displaystyle {\begin{aligned}&{{X}_{1,2}}=+sig\left(-B\right)\cdot {{y}_{1}}\pm \left({{y}_{2}}+{{y}_{3}}\right)\\&{{X}_{3,4}}=-sig\left(-B\right)\cdot {{y}_{1}}\pm \left({{y}_{2}}-{{y}_{3}}\right)\\\end{aligned}}}
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{\displaystyle a{{x}^{4}}+b{{x}^{3}}+c{{x}^{2}}+dx+e=0}
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{\displaystyle x=-{\frac {b}{4a}}+X}
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{\displaystyle {{X}^{4}}+\overbrace {\left(-{\frac {3{{b}^{2}}}{8{{a}^{2}}}}+{\frac {c}{a}}\right)} ^{A}\cdot {{X}^{2}}+\overbrace {\left({\frac {{b}^{3}}{8{{a}^{3}}}}-{\frac {bc}{2{{a}^{2}}}}+{\frac {d}{a}}\right)} ^{B}\cdot X+\overbrace {\left(-{\frac {3{{b}^{4}}}{256{{a}^{4}}}}+{\frac {{{b}^{2}}c}{16{{a}^{3}}}}-{\frac {bd}{4{{a}^{2}}}}+{\frac {e}{a}}\right)} ^{C}=0}
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{\displaystyle {{x}_{1,2,3,4}}=-{\frac {b}{4a}}+{{X}_{1,2,3,4}}}
Források Matematikai kisenciklopédia. szerk. Lukács Ernőné és Tarján Rezsőné. Budapest: Gondolat. 1968. 77-78. oldalKleine Enzyklopädie. Mathematik. Leipzig: VEB Verlag Enzyklopädie. 1970. 112-113. és 116. oldal.További információk 
![{\displaystyle u,v={\sqrt[{3}]{-{\frac {Q}{2}}\pm {\sqrt {\Delta }}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c6ee066f9be3662f902364792649e323e2b83d39)
![{\displaystyle {\begin{aligned}&\Delta ={{\left({\frac {Q}{2}}\right)}^{2}}+{{\left({\frac {P}{3}}\right)}^{3}}\\&u,v={\sqrt[{3}]{-{\frac {Q}{2}}\pm {\sqrt {\Delta }}}}\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3997427636e6fd3d417a0dfd2153dd5a270ded9d)
