(40+2x)(20+2x)-800=700

Solution for (40+2x)(20+2x)-800=700 equation:

(40+2x)(20+2x)-800=700

We move all terms to the left:

(40+2x)(20+2x)-800-(700)=0

We add all the numbers together, and all the variables

(2x+40)(2x+20)-800-700=0

We add all the numbers together, and all the variables

(2x+40)(2x+20)-1500=0

We multiply parentheses ..

(+4x^2+40x+80x+800)-1500=0

We get rid of parentheses

4x^2+40x+80x+800-1500=0

We add all the numbers together, and all the variables

4x^2+120x-700=0

a = 4; b = 120; c = -700;
Δ = b2-4ac
Δ = 1202-4·4·(-700)
Δ = 25600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25600}=160$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(120)-160}{2*4}=\frac{-280}{8}
=-35
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(120)+160}{2*4}=\frac{40}{8}
=5
$