x(x-3)+x(x-4)=12-x

Solution for x(x-3)+x(x-4)=12-x equation:

x(x-3)+x(x-4)=12-x

We move all terms to the left:

x(x-3)+x(x-4)-(12-x)=0

We add all the numbers together, and all the variables

x(x-3)+x(x-4)-(-1x+12)=0

We multiply parentheses

x^2+x^2-3x-4x-(-1x+12)=0

We get rid of parentheses

x^2+x^2-3x-4x+1x-12=0

We add all the numbers together, and all the variables

2x^2-6x-12=0

a = 2; b = -6; c = -12;
Δ = b2-4ac
Δ = -62-4·2·(-12)
Δ = 132
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{132}=\sqrt{4*33}=\sqrt{4}*\sqrt{33}=2\sqrt{33}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{33}}{2*2}=\frac{6-2\sqrt{33}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{33}}{2*2}=\frac{6+2\sqrt{33}}{4}
$