(20-2x)(10-2x)=96

Solution for (20-2x)(10-2x)=96 equation:

(20-2x)(10-2x)=96

We move all terms to the left:

(20-2x)(10-2x)-(96)=0

We add all the numbers together, and all the variables

(-2x+20)(-2x+10)-96=0

We multiply parentheses ..

(+4x^2-20x-40x+200)-96=0

We get rid of parentheses

4x^2-20x-40x+200-96=0

We add all the numbers together, and all the variables

4x^2-60x+104=0

a = 4; b = -60; c = +104;
Δ = b2-4ac
Δ = -602-4·4·104
Δ = 1936
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1936}=44$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-60)-44}{2*4}=\frac{16}{8}
=2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-60)+44}{2*4}=\frac{104}{8}
=13
$