(20+2x)(15+2x)-600=0

Solution for (20+2x)(15+2x)-600=0 equation:

(20+2x)(15+2x)-600=0

We add all the numbers together, and all the variables

(2x+20)(2x+15)-600=0

We multiply parentheses ..

(+4x^2+30x+40x+300)-600=0

We get rid of parentheses

4x^2+30x+40x+300-600=0

We add all the numbers together, and all the variables

4x^2+70x-300=0

a = 4; b = 70; c = -300;
Δ = b2-4ac
Δ = 702-4·4·(-300)
Δ = 9700
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:

$\sqrt{\Delta}=\sqrt{9700}=\sqrt{100*97}=\sqrt{100}*\sqrt{97}=10\sqrt{97}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(70)-10\sqrt{97}}{2*4}=\frac{-70-10\sqrt{97}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(70)+10\sqrt{97}}{2*4}=\frac{-70+10\sqrt{97}}{8}
$