(8x+2x)(20+2x)=540

Solution for (8x+2x)(20+2x)=540 equation:

(8x+2x)(20+2x)=540

We move all terms to the left:

(8x+2x)(20+2x)-(540)=0

We add all the numbers together, and all the variables

(+10x)(2x+20)-540=0

We multiply parentheses ..

(+20x^2+200x)-540=0

We get rid of parentheses

20x^2+200x-540=0

a = 20; b = 200; c = -540;
Δ = b2-4ac
Δ = 2002-4·20·(-540)
Δ = 83200
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:

$\sqrt{\Delta}=\sqrt{83200}=\sqrt{6400*13}=\sqrt{6400}*\sqrt{13}=80\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(200)-80\sqrt{13}}{2*20}=\frac{-200-80\sqrt{13}}{40}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(200)+80\sqrt{13}}{2*20}=\frac{-200+80\sqrt{13}}{40}
$