(x-5)(3x-4)-(2x+8)(x-5)=0

Solution for (x-5)(3x-4)-(2x+8)(x-5)=0 equation:

(x-5)(3x-4)-(2x+8)(x-5)=0

We multiply parentheses ..

(+3x^2-4x-15x+20)-(2x+8)(x-5)=0

We get rid of parentheses

3x^2-4x-15x-(2x+8)(x-5)+20=0

We multiply parentheses ..

3x^2-(+2x^2-10x+8x-40)-4x-15x+20=0

We add all the numbers together, and all the variables

3x^2-(+2x^2-10x+8x-40)-19x+20=0

We get rid of parentheses

3x^2-2x^2+10x-8x-19x+40+20=0

We add all the numbers together, and all the variables

x^2-17x+60=0

a = 1; b = -17; c = +60;
Δ = b2-4ac
Δ = -172-4·1·60
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-7}{2*1}=\frac{10}{2}
=5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+7}{2*1}=\frac{24}{2}
=12
$