(20+2x)(30+2x)-600=336

Solution for (20+2x)(30+2x)-600=336 equation:

(20+2x)(30+2x)-600=336

We move all terms to the left:

(20+2x)(30+2x)-600-(336)=0

We add all the numbers together, and all the variables

(2x+20)(2x+30)-600-336=0

We add all the numbers together, and all the variables

(2x+20)(2x+30)-936=0

We multiply parentheses ..

(+4x^2+60x+40x+600)-936=0

We get rid of parentheses

4x^2+60x+40x+600-936=0

We add all the numbers together, and all the variables

4x^2+100x-336=0

a = 4; b = 100; c = -336;
Δ = b2-4ac
Δ = 1002-4·4·(-336)
Δ = 15376
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{15376}=124$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(100)-124}{2*4}=\frac{-224}{8}
=-28
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(100)+124}{2*4}=\frac{24}{8}
=3
$