4x^2-8x-6x+12=x^2-x

Solution for 4x^2-8x-6x+12=x^2-x equation:

4x^2-8x-6x+12=x^2-x

We move all terms to the left:

4x^2-8x-6x+12-(x^2-x)=0

We add all the numbers together, and all the variables

4x^2-14x-(x^2-x)+12=0

We get rid of parentheses

4x^2-x^2-14x+x+12=0

We add all the numbers together, and all the variables

3x^2-13x+12=0

a = 3; b = -13; c = +12;
Δ = b2-4ac
Δ = -132-4·3·12
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-5}{2*3}=\frac{8}{6}
=1+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+5}{2*3}=\frac{18}{6}
=3
$