3x+12+6x-24=x^2+x

Solution for 3x+12+6x-24=x^2+x equation:

3x+12+6x-24=x^2+x

We move all terms to the left:

3x+12+6x-24-(x^2+x)=0

We add all the numbers together, and all the variables

9x-(x^2+x)-12=0

We get rid of parentheses

-x^2+9x-x-12=0

We add all the numbers together, and all the variables

-1x^2+8x-12=0

a = -1; b = 8; c = -12;
Δ = b2-4ac
Δ = 82-4·(-1)·(-12)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4}{2*-1}=\frac{-12}{-2}
=+6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4}{2*-1}=\frac{-4}{-2}
=+2
$