(12-2x)(20-2x)=93

Solution for (12-2x)(20-2x)=93 equation:

(12-2x)(20-2x)=93

We move all terms to the left:

(12-2x)(20-2x)-(93)=0

We add all the numbers together, and all the variables

(-2x+12)(-2x+20)-93=0

We multiply parentheses ..

(+4x^2-40x-24x+240)-93=0

We get rid of parentheses

4x^2-40x-24x+240-93=0

We add all the numbers together, and all the variables

4x^2-64x+147=0

a = 4; b = -64; c = +147;
Δ = b2-4ac
Δ = -642-4·4·147
Δ = 1744
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:

$\sqrt{\Delta}=\sqrt{1744}=\sqrt{16*109}=\sqrt{16}*\sqrt{109}=4\sqrt{109}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-64)-4\sqrt{109}}{2*4}=\frac{64-4\sqrt{109}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-64)+4\sqrt{109}}{2*4}=\frac{64+4\sqrt{109}}{8}
$