(30+2x)(20+2x)=1200

Solution for (30+2x)(20+2x)=1200 equation:

(30+2x)(20+2x)=1200

We move all terms to the left:

(30+2x)(20+2x)-(1200)=0

We add all the numbers together, and all the variables

(2x+30)(2x+20)-1200=0

We multiply parentheses ..

(+4x^2+40x+60x+600)-1200=0

We get rid of parentheses

4x^2+40x+60x+600-1200=0

We add all the numbers together, and all the variables

4x^2+100x-600=0

a = 4; b = 100; c = -600;
Δ = b2-4ac
Δ = 1002-4·4·(-600)
Δ = 19600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{19600}=140$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(100)-140}{2*4}=\frac{-240}{8}
=-30
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(100)+140}{2*4}=\frac{40}{8}
=5
$