(4x-5)(2x-2)=(3x)(3x+3)

Solution for (4x-5)(2x-2)=(3x)(3x+3) equation:

(4x-5)(2x-2)=(3x)(3x+3)

We move all terms to the left:

(4x-5)(2x-2)-((3x)(3x+3))=0

We multiply parentheses ..

(+8x^2-8x-10x+10)-(3x(3x+3))=0


We calculate terms in parentheses: -(3x(3x+3)), so:

3x(3x+3)

We multiply parentheses

9x^2+9x

Back to the equation:

-(9x^2+9x)


We get rid of parentheses

8x^2-9x^2-8x-10x-9x+10=0

We add all the numbers together, and all the variables

-1x^2-27x+10=0

a = -1; b = -27; c = +10;
Δ = b2-4ac
Δ = -272-4·(-1)·10
Δ = 769
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-27)-\sqrt{769}}{2*-1}=\frac{27-\sqrt{769}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-27)+\sqrt{769}}{2*-1}=\frac{27+\sqrt{769}}{-2}
$