((2x-2)(x+5))=180

Solution for ((2x-2)(x+5))=180 equation:

((2x-2)(x+5))=180

We move all terms to the left:

((2x-2)(x+5))-(180)=0

We multiply parentheses ..

((+2x^2+10x-2x-10))-180=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-10-180=0

We add all the numbers together, and all the variables

2x^2+8x-190=0

a = 2; b = 8; c = -190;
Δ = b2-4ac
Δ = 82-4·2·(-190)
Δ = 1584
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:

$\sqrt{\Delta}=\sqrt{1584}=\sqrt{144*11}=\sqrt{144}*\sqrt{11}=12\sqrt{11}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-12\sqrt{11}}{2*2}=\frac{-8-12\sqrt{11}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+12\sqrt{11}}{2*2}=\frac{-8+12\sqrt{11}}{4}
$