(18+2x)(20+2x)=1368

Solution for (18+2x)(20+2x)=1368 equation:

(18+2x)(20+2x)=1368

We move all terms to the left:

(18+2x)(20+2x)-(1368)=0

We add all the numbers together, and all the variables

(2x+18)(2x+20)-1368=0

We multiply parentheses ..

(+4x^2+40x+36x+360)-1368=0

We get rid of parentheses

4x^2+40x+36x+360-1368=0

We add all the numbers together, and all the variables

4x^2+76x-1008=0

a = 4; b = 76; c = -1008;
Δ = b2-4ac
Δ = 762-4·4·(-1008)
Δ = 21904
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{21904}=148$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(76)-148}{2*4}=\frac{-224}{8}
=-28
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(76)+148}{2*4}=\frac{72}{8}
=9
$