(14-2x)(20-2x)=128

Solution for (14-2x)(20-2x)=128 equation:

(14-2x)(20-2x)=128

We move all terms to the left:

(14-2x)(20-2x)-(128)=0

We add all the numbers together, and all the variables

(-2x+14)(-2x+20)-128=0

We multiply parentheses ..

(+4x^2-40x-28x+280)-128=0

We get rid of parentheses

4x^2-40x-28x+280-128=0

We add all the numbers together, and all the variables

4x^2-68x+152=0

a = 4; b = -68; c = +152;
Δ = b2-4ac
Δ = -682-4·4·152
Δ = 2192
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:

$\sqrt{\Delta}=\sqrt{2192}=\sqrt{16*137}=\sqrt{16}*\sqrt{137}=4\sqrt{137}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-68)-4\sqrt{137}}{2*4}=\frac{68-4\sqrt{137}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-68)+4\sqrt{137}}{2*4}=\frac{68+4\sqrt{137}}{8}
$