(12+x)(18+x)=432

Solution for (12+x)(18+x)=432 equation:

(12+x)(18+x)=432

We move all terms to the left:

(12+x)(18+x)-(432)=0

We add all the numbers together, and all the variables

(x+12)(x+18)-432=0

We multiply parentheses ..

(+x^2+18x+12x+216)-432=0

We get rid of parentheses

x^2+18x+12x+216-432=0

We add all the numbers together, and all the variables

x^2+30x-216=0

a = 1; b = 30; c = -216;
Δ = b2-4ac
Δ = 302-4·1·(-216)
Δ = 1764
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1764}=42$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(30)-42}{2*1}=\frac{-72}{2}
=-36
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(30)+42}{2*1}=\frac{12}{2}
=6
$