(15+2X)(20+2x)=336

Solution for (15+2X)(20+2x)=336 equation:

(15+2X)(20+2X)=336

We move all terms to the left:

(15+2X)(20+2X)-(336)=0

We add all the numbers together, and all the variables

(2X+15)(2X+20)-336=0

We multiply parentheses ..

(+4X^2+40X+30X+300)-336=0

We get rid of parentheses

4X^2+40X+30X+300-336=0

We add all the numbers together, and all the variables

4X^2+70X-36=0

a = 4; b = 70; c = -36;
Δ = b2-4ac
Δ = 702-4·4·(-36)
Δ = 5476
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{5476}=74$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(70)-74}{2*4}=\frac{-144}{8}
=-18
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(70)+74}{2*4}=\frac{4}{8}
=1/2
$