(X+4)(x+5)=132

Solution for (X+4)(x+5)=132 equation:

(X+4)(X+5)=132

We move all terms to the left:

(X+4)(X+5)-(132)=0

We multiply parentheses ..

(+X^2+5X+4X+20)-132=0

We get rid of parentheses

X^2+5X+4X+20-132=0

We add all the numbers together, and all the variables

X^2+9X-112=0

a = 1; b = 9; c = -112;
Δ = b2-4ac
Δ = 92-4·1·(-112)
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-23}{2*1}=\frac{-32}{2}
=-16
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+23}{2*1}=\frac{14}{2}
=7
$