(15-2x)(10-2x)=120

Solution for (15-2x)(10-2x)=120 equation:

(15-2x)(10-2x)=120

We move all terms to the left:

(15-2x)(10-2x)-(120)=0

We add all the numbers together, and all the variables

(-2x+15)(-2x+10)-120=0

We multiply parentheses ..

(+4x^2-20x-30x+150)-120=0

We get rid of parentheses

4x^2-20x-30x+150-120=0

We add all the numbers together, and all the variables

4x^2-50x+30=0

a = 4; b = -50; c = +30;
Δ = b2-4ac
Δ = -502-4·4·30
Δ = 2020
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:

$\sqrt{\Delta}=\sqrt{2020}=\sqrt{4*505}=\sqrt{4}*\sqrt{505}=2\sqrt{505}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-50)-2\sqrt{505}}{2*4}=\frac{50-2\sqrt{505}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-50)+2\sqrt{505}}{2*4}=\frac{50+2\sqrt{505}}{8}
$