(10-2x)(12-2x)=21

Solution for (10-2x)(12-2x)=21 equation:

(10-2x)(12-2x)=21

We move all terms to the left:

(10-2x)(12-2x)-(21)=0

We add all the numbers together, and all the variables

(-2x+10)(-2x+12)-21=0

We multiply parentheses ..

(+4x^2-24x-20x+120)-21=0

We get rid of parentheses

4x^2-24x-20x+120-21=0

We add all the numbers together, and all the variables

4x^2-44x+99=0

a = 4; b = -44; c = +99;
Δ = b2-4ac
Δ = -442-4·4·99
Δ = 352
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:

$\sqrt{\Delta}=\sqrt{352}=\sqrt{16*22}=\sqrt{16}*\sqrt{22}=4\sqrt{22}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-44)-4\sqrt{22}}{2*4}=\frac{44-4\sqrt{22}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-44)+4\sqrt{22}}{2*4}=\frac{44+4\sqrt{22}}{8}
$