(5-x)(3x-1)-(4x+2)(5-x)=0

Solution for (5-x)(3x-1)-(4x+2)(5-x)=0 equation:

(5-x)(3x-1)-(4x+2)(5-x)=0

We add all the numbers together, and all the variables

(-1x+5)(3x-1)-(4x+2)(-1x+5)=0

We multiply parentheses ..

(-3x^2+x+15x-5)-(4x+2)(-1x+5)=0

We get rid of parentheses

-3x^2+x+15x-(4x+2)(-1x+5)-5=0

We multiply parentheses ..

-3x^2-(-4x^2+20x-2x+10)+x+15x-5=0

We add all the numbers together, and all the variables

-3x^2-(-4x^2+20x-2x+10)+16x-5=0

We get rid of parentheses

-3x^2+4x^2-20x+2x+16x-10-5=0

We add all the numbers together, and all the variables

x^2-2x-15=0

a = 1; b = -2; c = -15;
Δ = b2-4ac
Δ = -22-4·1·(-15)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-8}{2*1}=\frac{-6}{2}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+8}{2*1}=\frac{10}{2}
=5
$