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

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

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

We move all terms to the left:

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

We multiply parentheses

24x+12x-(3(x^2-9))+36-36=0


We calculate terms in parentheses: -(3(x^2-9)), so:

3(x^2-9)

We multiply parentheses

3x^2-27

Back to the equation:

-(3x^2-27)


We add all the numbers together, and all the variables

36x-(3x^2-27)=0

We get rid of parentheses

-3x^2+36x+27=0

a = -3; b = 36; c = +27;
Δ = b2-4ac
Δ = 362-4·(-3)·27
Δ = 1620
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:

$\sqrt{\Delta}=\sqrt{1620}=\sqrt{324*5}=\sqrt{324}*\sqrt{5}=18\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-18\sqrt{5}}{2*-3}=\frac{-36-18\sqrt{5}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+18\sqrt{5}}{2*-3}=\frac{-36+18\sqrt{5}}{-6}
$