X^2+(3-2i)*x+5-5i=0.... X is a complex number.... find the roots, can anyone help me with detailed solution

It's a quadratic; it can be solved with the usual formula.
Only trick is finding a complex square root. Use DeMoivre's Theorem, along with the half-angle trig identities for that.

With a = 1:
x^2 + bx + c = 0

that becomes
x = (-b ± √(b^2 - 4c))/2 = -b/2 ± √((b/2)^2 - c)

With b = 3 - 2i, c = 5 - 5i
b^2 = 3^2 - 12i + 4(i^2) = 5 - 12i
Let z = b^2 - 4c = -15 + 8i
= r cisθ = r(cosθ + i sinθ)

Then ±√z = (√r) cis({θ, θ+2π} /2) = (√r) cis{θ/2, θ/2 + π},

r^2 = 15^2 + 8^2 = 289 = 17^2; r = 17
cosθ = -15/17, sinθ = 8/17
cos(θ/2) = ±√[(1 + cosθ)/2] = ±√[(17 - 15)/(2*17)] = ±1/√17
sin(θ/2) = ±√[(1 - cosθ)/2] = ±√[(17 + 15)/(2*17)] = ±4/√17
with the signs matching each other, both +, or both -
(θ is in Quadrant 2, so θ/2 is in Q1; θ/2 + π is in Q3)

√z = ±√17 (1/√17 + 4i/√17) = ±(1 + 4i)

x = (-3 + 2i ± (1 + 4i))/2 = {-1 + 3i, -2 - i}

I will leave it to you to check that they both satisfy the original quadratic.

x^2 + (3-2i)x + 5-5i = 0

x = (-3+2i ± √((-3+2i)^2 - 4(5-5i)) / 2

x = (-3+2i ± √(5 - 12i - 20 + 20i) / 2

x = (-3+2i ± √(-15 + 8i) / 2

= = = = =
You can solve for √(-15 + 8i) without using trigonometry

Assume there are integers a and b such that √(-15 + 8i) = a + bi

-15 + 8i = (a^2 - b^2) + 2abi

b^2 - a^2 = 15

ab = 4
a^2 b^2 = 16

b^4 - a^2 b^2 = 15 b^2

b^4 - 15 b^2 - 16 = 0

(b^2 + 1)(b^2 - 16) = 0

b = 4 or -4
a = 1 or -1

√(-15 + 8i) = ±(1 + 4i)

= = = = =

x = (-3+2i ± (1 + 4i) / 2

x = -1 + 3i, -2 - i

check

(-1 + 3i)^2 + (3 - 2i)(-1 + 3i) + 5-5i =
-8 - 6i + 3 + 11i + 5 - 5i =
0

(-2 - i)^2 + (3 - 2i)(-2 - i) + 5-5i =
3 + 4i - 8 + i + 5 - 5i =
0

x = -(3-2i) +-√(9-12i-4)-4(5-5i) / 2 This is the quadratic equation solution

x= -(3-2i)/2 +- √(5-12i-20+20i) / 2 = -(3-2i)/2 +- √(-15+8i) / 2