(20+2x)(12+2x)-240=560

Solution for (20+2x)(12+2x)-240=560 equation:

(20+2x)(12+2x)-240=560

We move all terms to the left:

(20+2x)(12+2x)-240-(560)=0

We add all the numbers together, and all the variables

(2x+20)(2x+12)-240-560=0

We add all the numbers together, and all the variables

(2x+20)(2x+12)-800=0

We multiply parentheses ..

(+4x^2+24x+40x+240)-800=0

We get rid of parentheses

4x^2+24x+40x+240-800=0

We add all the numbers together, and all the variables

4x^2+64x-560=0

a = 4; b = 64; c = -560;
Δ = b2-4ac
Δ = 642-4·4·(-560)
Δ = 13056
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:

$\sqrt{\Delta}=\sqrt{13056}=\sqrt{256*51}=\sqrt{256}*\sqrt{51}=16\sqrt{51}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(64)-16\sqrt{51}}{2*4}=\frac{-64-16\sqrt{51}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(64)+16\sqrt{51}}{2*4}=\frac{-64+16\sqrt{51}}{8}
$