(2x+40)*(2x+20)=700

Solution for (2x+40)*(2x+20)=700 equation:

(2x+40)(2x+20)=700

We move all terms to the left:

(2x+40)(2x+20)-(700)=0

We multiply parentheses ..

(+4x^2+40x+80x+800)-700=0

We get rid of parentheses

4x^2+40x+80x+800-700=0

We add all the numbers together, and all the variables

4x^2+120x+100=0

a = 4; b = 120; c = +100;
Δ = b2-4ac
Δ = 1202-4·4·100
Δ = 12800
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:

$\sqrt{\Delta}=\sqrt{12800}=\sqrt{6400*2}=\sqrt{6400}*\sqrt{2}=80\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(120)-80\sqrt{2}}{2*4}=\frac{-120-80\sqrt{2}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(120)+80\sqrt{2}}{2*4}=\frac{-120+80\sqrt{2}}{8}
$