(x+12)(x+16)=320

Solution for (x+12)(x+16)=320 equation:

(x+12)(x+16)=320

We move all terms to the left:

(x+12)(x+16)-(320)=0

We multiply parentheses ..

(+x^2+16x+12x+192)-320=0

We get rid of parentheses

x^2+16x+12x+192-320=0

We add all the numbers together, and all the variables

x^2+28x-128=0

a = 1; b = 28; c = -128;
Δ = b2-4ac
Δ = 282-4·1·(-128)
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-36}{2*1}=\frac{-64}{2}
=-32
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+36}{2*1}=\frac{8}{2}
=4
$