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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-2x^2-8x+5x+20)=0

We get rid of parentheses

-2x^2-8x+5x+20=0

We add all the numbers together, and all the variables

-2x^2-3x+20=0

a = -2; b = -3; c = +20;
Δ = b2-4ac
Δ = -32-4·(-2)·20
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-13}{2*-2}=\frac{-10}{-4}
=2+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+13}{2*-2}=\frac{16}{-4}
=-4
$