9x^2-3(x^2+x)+4x-3+2x-5=12

Solution for 9x^2-3(x^2+x)+4x-3+2x-5=12 equation:

9x^2-3(x^2+x)+4x-3+2x-5=12

We move all terms to the left:

9x^2-3(x^2+x)+4x-3+2x-5-(12)=0

We add all the numbers together, and all the variables

9x^2+6x-3(x^2+x)-20=0

We multiply parentheses

9x^2-3x^2+6x-3x-20=0

We add all the numbers together, and all the variables

6x^2+3x-20=0

a = 6; b = 3; c = -20;
Δ = b2-4ac
Δ = 32-4·6·(-20)
Δ = 489
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{-(3)-\sqrt{489}}{2*6}=\frac{-3-\sqrt{489}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{489}}{2*6}=\frac{-3+\sqrt{489}}{12}
$