(10x-5)(2x+4)=0

Solution for (10x-5)(2x+4)=0 equation:

(10x-5)(2x+4)=0

We multiply parentheses ..

(+20x^2+40x-10x-20)=0

We get rid of parentheses

20x^2+40x-10x-20=0

We add all the numbers together, and all the variables

20x^2+30x-20=0

a = 20; b = 30; c = -20;
Δ = b2-4ac
Δ = 302-4·20·(-20)
Δ = 2500
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2500}=50$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(30)-50}{2*20}=\frac{-80}{40}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(30)+50}{2*20}=\frac{20}{40}
=1/2
$