This is the method for solving any quartic equation.
Say we have:
ax⁴+bx³+cx²+dx+e=0 .
Divide by a , to rewrite this as:
x⁴+fx³+gx²+hx+j=0 .
Put y=x+f/4 , so x=y−f/4 , so

 

x4 = y4−f3y+3f2y2/8−f3y/16+f4/256 x3 = y3−3y2f/4+3yf2/16−f3/64 x2 = y2−2yf/4+f2/16

so x⁴+fx³+gx²+hx+j=y⁴−fy³+3f²/y²/8−f³y/16+f⁴/256+fy³−3y²f²/4+3yf³/16−f⁴/64+gy²−gyf/2+gf²/16+hy−hf/4+j
=y⁴+y³(−f+f) y²(3f²/8)−3²/4+4) +y(−f³/16+3f³/16−gf/2+h) +

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f⁴/256−f⁴/64+gf²/16−hf/4+j

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So we can rewrite the original equation as:
y⁴+py²+qy+r=0
with p=−3f²/8+g ,
q=f³/8−gf/2+h ,
r=−3f⁴/256+gf²/16−hf/4+j .
Now, we can write:
y⁴+py²=−qy−r
y⁴+2py²+p²=py²−qy−r+p²
(y²+p)²=py²−qy−r+p² .
Now, for any z ,
(y²+p+z)²=((y²+p)+z)²
=(y²+p)²+2(y²+p)z+z²
=py²−qy−r+p²+2z(y²+p)+z²
=(p+2z)y²−qy+(p²−r+2pz+z²) (*)
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)(p²−r+2pz+z²)=0 .
We can rewrite this as (q²−4p³+4pr)+(−16p²+8r)z−20pz²−8z³=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.