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

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

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

We move all terms to the left:

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

We multiply parentheses

-3x^2+(x-4)+(2x-5)+12x-8=0

We get rid of parentheses

-3x^2+x+2x+12x-4-5-8=0

We add all the numbers together, and all the variables

-3x^2+15x-17=0

a = -3; b = 15; c = -17;
Δ = b2-4ac
Δ = 152-4·(-3)·(-17)
Δ = 21
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-\sqrt{21}}{2*-3}=\frac{-15-\sqrt{21}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+\sqrt{21}}{2*-3}=\frac{-15+\sqrt{21}}{-6}
$