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

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

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

We add all the numbers together, and all the variables

(-1x+5)(4x+10)=0

We multiply parentheses ..

(-4x^2-10x+20x+50)=0

We get rid of parentheses

-4x^2-10x+20x+50=0

We add all the numbers together, and all the variables

-4x^2+10x+50=0

a = -4; b = 10; c = +50;
Δ = b2-4ac
Δ = 102-4·(-4)·50
Δ = 900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{900}=30$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-30}{2*-4}=\frac{-40}{-8}
=+5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+30}{2*-4}=\frac{20}{-8}
=-2+1/2
$