Solve the equation x^3 - 6x^2 + 13x + 10 = 0 if the roots are in A.P

Find the roots

there will be 3 roots to the equation
so let the roots be
a, a+d, a + 2d
then

(x-a)(x - a - d)(x - a - 2d) = x^3 - 6x^2 + 13x + 10
(x^2 - ax - dx - ax + a^2 + ad)(x - a - 2d) = x^3 - 6x^2 + 13x + 10
(x^2 - 2ax - dx + a^2 + ad)(x - a - 2d) = x^3 - 6x^2 + 13x + 10
x^3 - ax^2 - 2dx^2 - 2ax^2 + 2a^2x + 4 adx - dx^2 + adx + 2d^2x + a^2x - a^3 - 2a^2d + adx - a^2d - 2ad^2 = x^3 - 6x^2 + 13x + 10

x^3 - 3 ax^2 - 3dx^2 + 3a^2x + 6adx + 2d^2x - a^3 - 3 a^2d - 2ad^2 = x^3 - 6x^2 + 13x + 10

x^3 + x^2(-3a - 3d) + x (3 a^2 + 6ad + 2d^2) - a^3 - 3a^2d - 2ad^2 = x^3 - 6x^2 + 13x + 10
3a + 3d = 6
3 a^2 + 6ad + 2d^2 = 13
- a^3 - 3a^2d - 2ad^2 = 10
on solving these 3 equations
we get that no solution exists!

so if coefficient of x^2 was 6
the solution was
a = - 2 - i
d = i

roots are
- 2 - i, -2, -2 + i

No.