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

Solution for (20-2x)(10-2x)=0 equation:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

4x^2-60x+200=0

a = 4; b = -60; c = +200;
Δ = b2-4ac
Δ = -602-4·4·200
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-60)-20}{2*4}=\frac{40}{8}
=5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-60)+20}{2*4}=\frac{80}{8}
=10
$