(8+2x)*(10+2x)=160

Solution for (8+2x)*(10+2x)=160 equation:

(8+2x)(10+2x)=160

We move all terms to the left:

(8+2x)(10+2x)-(160)=0

We add all the numbers together, and all the variables

(2x+8)(2x+10)-160=0

We multiply parentheses ..

(+4x^2+20x+16x+80)-160=0

We get rid of parentheses

4x^2+20x+16x+80-160=0

We add all the numbers together, and all the variables

4x^2+36x-80=0

a = 4; b = 36; c = -80;
Δ = b2-4ac
Δ = 362-4·4·(-80)
Δ = 2576
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:

$\sqrt{\Delta}=\sqrt{2576}=\sqrt{16*161}=\sqrt{16}*\sqrt{161}=4\sqrt{161}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-4\sqrt{161}}{2*4}=\frac{-36-4\sqrt{161}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+4\sqrt{161}}{2*4}=\frac{-36+4\sqrt{161}}{8}
$