x(3x-6)+2x(x-3)=36+x^2

Solution for x(3x-6)+2x(x-3)=36+x^2 equation:

x(3x-6)+2x(x-3)=36+x^2

We move all terms to the left:

x(3x-6)+2x(x-3)-(36+x^2)=0

We multiply parentheses

-(36+x^2)+3x^2+2x^2-6x-6x=0

We get rid of parentheses

-x^2+3x^2+2x^2-6x-6x-36=0

We add all the numbers together, and all the variables

4x^2-12x-36=0

a = 4; b = -12; c = -36;
Δ = b2-4ac
Δ = -122-4·4·(-36)
Δ = 720
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:

$\sqrt{\Delta}=\sqrt{720}=\sqrt{144*5}=\sqrt{144}*\sqrt{5}=12\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-12\sqrt{5}}{2*4}=\frac{12-12\sqrt{5}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+12\sqrt{5}}{2*4}=\frac{12+12\sqrt{5}}{8}
$