(x-12)(x+28)=(x+4)(x-4)

Solution for (x-12)(x+28)=(x+4)(x-4) equation:

(x-12)(x+28)=(x+4)(x-4)

We move all terms to the left:

(x-12)(x+28)-((x+4)(x-4))=0

We use the square of the difference formula

x^2+(x-12)(x+28)+16=0

We multiply parentheses ..

x^2+(+x^2+28x-12x-336)+16=0

We get rid of parentheses

x^2+x^2+28x-12x-336+16=0

We add all the numbers together, and all the variables

2x^2+16x-320=0

a = 2; b = 16; c = -320;
Δ = b2-4ac
Δ = 162-4·2·(-320)
Δ = 2816
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2816}=\sqrt{256*11}=\sqrt{256}*\sqrt{11}=16\sqrt{11}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-16\sqrt{11}}{2*2}=\frac{-16-16\sqrt{11}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16\sqrt{11}}{2*2}=\frac{-16+16\sqrt{11}}{4}
$