90=(2x-2)(x+5)

Solution for 90=(2x-2)(x+5) equation:

90=(2x-2)(x+5)

We move all terms to the left:

90-((2x-2)(x+5))=0

We multiply parentheses ..

-((+2x^2+10x-2x-10))+90=0


We calculate terms in parentheses: -((+2x^2+10x-2x-10)), so:

(+2x^2+10x-2x-10)

We get rid of parentheses

2x^2+10x-2x-10

We add all the numbers together, and all the variables

2x^2+8x-10

Back to the equation:

-(2x^2+8x-10)


We get rid of parentheses

-2x^2-8x+10+90=0

We add all the numbers together, and all the variables

-2x^2-8x+100=0

a = -2; b = -8; c = +100;
Δ = b2-4ac
Δ = -82-4·(-2)·100
Δ = 864
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{864}=\sqrt{144*6}=\sqrt{144}*\sqrt{6}=12\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-12\sqrt{6}}{2*-2}=\frac{8-12\sqrt{6}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+12\sqrt{6}}{2*-2}=\frac{8+12\sqrt{6}}{-4}
$