Descartes Solution
A quartic equation has the form
x ^{4} + bx ^{3} + cx ^{2} + dx + e = 0
I will illustrate the steps by which we can arrive at the solution of a biquadratic or quartic equation.
 Express x^{4} + bx^{3} + cx^{2} + dx + e = 0 as
x^{4} + px^{2} + qx + r = 0.
 Let x^{4} + px^{2} + qx + r = 0 be equivalent to
x^{4} + px^{2} + qx + r = (x^{2} + Ax + B)(x^{2} + Cx + D)
 You will get A = C , B + D + AC = p, AD + BC = q and BD = r
 B + D  A^{2} = p, A(D  B) = q and BD = r
From these frame a cubic equation in A.
(B + D) = (p + A^{2})
D  B = q/A
DB = r
(B+D)^{2} = (DB)^{2} + 4DB
(p + A^{2})^{2} = (q/A)^{2} + 4r
(pA + A^{3})^{2} = q^{2} + 4rA^{2}
A^{6} + 2pA^{4} + p^{2}A^{2} = q^{2} + 4rA^{2}
A^{6} + 2pA^{4} + (p^{2}4r)A^{2}  q^{2} = 0
 Solve the cubic equation A^{6} + 2pA^{4} + (p^{2}4r)A^{2}  q^{2} = 0
You will get one real root.
 Find the value of D and B with the help of A.
 Substitute the value of A, C, B and D in the equations
(x^{2} + Ax + B) = 0
and (x^{2} + Cx + D)= 0 to get two quadratic equations.
 Solve the quadratic equations to get the roots.
Ferrari's solution
A quartic equation has the form
x ^{4} + bx ^{3} + cx ^{2} + dx + e = 0
I will illustrate the steps by which we can arrive at the solution of a biquadratic or quartic equation.
 Add (px + q)^{2} to both sides of the equation
x^{4} + bx^{3} + cx^{2} + dx + e = 0
x^{4} + bx^{3} + (c+p^{2})x^{2} + (d+2pq)x + (e+q^{2}) = (px + q)^{2}
 Choose p and q such that both sides of the equation is a perfect square.
(x^{2} + (b/2)x + k)^{2} = (px + q)^{2}
 Equate the constants of
x^{4} + bx^{3} + (c+p^{2})x^{2} + (d+2pq)x + (e+q^{2}) = (px + q)^{2}
and
(x^{2} + (b/2)x + k)^{2} = (px + q)^{2}
to get a cubic equation in k. This cubic equation is known as resolvent cubic.
 Find the values of p and q with the help of k_{1} = value of k.
 Get two equation from x^{2} + (b/2)x + k_{1} = ±(px + q)
x^{2} + (b/2 + p)x + k_{1}+q = 0
and x^{2} + (b/2  p)x + k_{1}q = 0
 Solve the two equations to get the solution.
