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

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

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

We multiply parentheses ..

(-2x^2+10x+4x-20)=0

We get rid of parentheses

-2x^2+10x+4x-20=0

We add all the numbers together, and all the variables

-2x^2+14x-20=0

a = -2; b = 14; c = -20;
Δ = b2-4ac
Δ = 142-4·(-2)·(-20)
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-6}{2*-2}=\frac{-20}{-4}
=+5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+6}{2*-2}=\frac{-8}{-4}
=+2
$