(2x+2)(x+2)=160

Solution for (2x+2)(x+2)=160 equation:

(2x+2)(x+2)=160

We move all terms to the left:

(2x+2)(x+2)-(160)=0

We multiply parentheses ..

(+2x^2+4x+2x+4)-160=0

We get rid of parentheses

2x^2+4x+2x+4-160=0

We add all the numbers together, and all the variables

2x^2+6x-156=0

a = 2; b = 6; c = -156;
Δ = b2-4ac
Δ = 62-4·2·(-156)
Δ = 1284
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:

$\sqrt{\Delta}=\sqrt{1284}=\sqrt{4*321}=\sqrt{4}*\sqrt{321}=2\sqrt{321}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{321}}{2*2}=\frac{-6-2\sqrt{321}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{321}}{2*2}=\frac{-6+2\sqrt{321}}{4}
$