This is the method for solving any quartic equation.
Say we have:
ax^{4}+bx^{3}+cx^{2}+dx+e=0 .
Divide by a , to rewrite this as:
x^{4}+fx^{3}+gx^{2}+hx+j=0 .
Put y=x+f/4 , so x=y−f/4 , so
so x^{4}+fx^{3}+gx^{2}+hx+j=y^{4}−fy^{3}+3f^{2}/y^{2}/8−f^{3}y/16+f^{4}/256+fy^{3}−3y^{2}f^{2}/4+3yf^{3}/16−f^{4}/64+gy^{2}−gyf/2+gf^{2}/16+hy−hf/4+j
=y^{4}+y^{3}(−f+f) y^{2}(3f^{2}/8)−3^{2}/4+4) +y(−f^{3}/16+3f^{3}/16−gf/2+h) +
f^{4}/256−f^{4}/64+gf^{2}/16−hf/4+j
So we can rewrite the original
equation as:
y^{4}+py^{2}+qy+r=0
with p=−3f^{2}/8+g ,
q=f^{3}/8−gf/2+h ,
r=−3f^{4}/256+gf^{2}/16−hf/4+j .
Now, we can write:
y^{4}+py^{2}=−qy−r
y^{4}+2py^{2}+p^{2}=py^{2}−qy−r+p^{2}
(y^{2}+p)^{2}=py^{2}−qy−r+p^{2} .
Now, for any z ,
(y^{2}+p+z)^{2}=((y^{2}+p)+z)^{2}
=(y^{2}+p)^{2}+2(y^{2}+p)z+z^{2}
=py^{2}−qy−r+p^{2}+2z(y^{2}+p)+z^{2}
=(p+2z)y^{2}−qy+(p^{2}−r+2pz+z^{2}) (*)
The right hand side of (*) is a quadratic in y ; and we can choose z so that it is a perfect square, i.e. so that the discriminant is zero, ie:
(−q)^{2}−4(p+2z)(p^{2}−r+2pz+z^{2})=0 .
We can rewrite this as (q^{2}−4p^{3}+4pr)+(−16p^{2}+8r)z−20pz^{2}−8z^{3}=0 .
This is a cubic equation in z . So we can solve it for z using the cubic equation formula (Cardano's method).
When we have solved this to find a value for z , we can substitute in this value of z to (*). This makes the right hand side of (*) a perfect square, so we can take the square root of both sides of (*). (*) is then a quadratic equation in y , which we can solve using the quadratic solution formula. The values of y will easily give us values of x , i.e. solutions of the original equation.
There can be 0, 1, 2, 3, or 4 real solutions.
Say we have:
ax^{4}+bx^{3}+cx^{2}+dx+e=0 .
Divide by a , to rewrite this as:
x^{4}+fx^{3}+gx^{2}+hx+j=0 .
Put y=x+f/4 , so x=y−f/4 , so

so x^{4}+fx^{3}+gx^{2}+hx+j=y^{4}−fy^{3}+3f^{2}/y^{2}/8−f^{3}y/16+f^{4}/256+fy^{3}−3y^{2}f^{2}/4+3yf^{3}/16−f^{4}/64+gy^{2}−gyf/2+gf^{2}/16+hy−hf/4+j
=y^{4}+y^{3}(−f+f) y^{2}(3f^{2}/8)−3^{2}/4+4) +y(−f^{3}/16+3f^{3}/16−gf/2+h) +
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y^{4}+py^{2}+qy+r=0
with p=−3f^{2}/8+g ,
q=f^{3}/8−gf/2+h ,
r=−3f^{4}/256+gf^{2}/16−hf/4+j .
Now, we can write:
y^{4}+py^{2}=−qy−r
y^{4}+2py^{2}+p^{2}=py^{2}−qy−r+p^{2}
(y^{2}+p)^{2}=py^{2}−qy−r+p^{2} .
Now, for any z ,
(y^{2}+p+z)^{2}=((y^{2}+p)+z)^{2}
=(y^{2}+p)^{2}+2(y^{2}+p)z+z^{2}
=py^{2}−qy−r+p^{2}+2z(y^{2}+p)+z^{2}
=(p+2z)y^{2}−qy+(p^{2}−r+2pz+z^{2}) (*)
The right hand side of (*) is a quadratic in y ; and we can choose z so that it is a perfect square, i.e. so that the discriminant is zero, ie:
(−q)^{2}−4(p+2z)(p^{2}−r+2pz+z^{2})=0 .
We can rewrite this as (q^{2}−4p^{3}+4pr)+(−16p^{2}+8r)z−20pz^{2}−8z^{3}=0 .
This is a cubic equation in z . So we can solve it for z using the cubic equation formula (Cardano's method).
When we have solved this to find a value for z , we can substitute in this value of z to (*). This makes the right hand side of (*) a perfect square, so we can take the square root of both sides of (*). (*) is then a quadratic equation in y , which we can solve using the quadratic solution formula. The values of y will easily give us values of x , i.e. solutions of the original equation.
There can be 0, 1, 2, 3, or 4 real solutions.