(3x-6)+(x-9)(x)=180

Solution for (3x-6)+(x-9)(x)=180 equation:

(3x-6)+(x-9)(x)=180

We move all terms to the left:

(3x-6)+(x-9)(x)-(180)=0

We multiply parentheses

x^2+(3x-6)-9x-180=0

We get rid of parentheses

x^2+3x-9x-6-180=0

We add all the numbers together, and all the variables

x^2-6x-186=0

a = 1; b = -6; c = -186;
Δ = b2-4ac
Δ = -62-4·1·(-186)
Δ = 780
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:

$\sqrt{\Delta}=\sqrt{780}=\sqrt{4*195}=\sqrt{4}*\sqrt{195}=2\sqrt{195}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{195}}{2*1}=\frac{6-2\sqrt{195}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{195}}{2*1}=\frac{6+2\sqrt{195}}{2}
$