x^2-3x+(x+3)(x-4)=180

Solution for x^2-3x+(x+3)(x-4)=180 equation:

x^2-3x+(x+3)(x-4)=180

We move all terms to the left:

x^2-3x+(x+3)(x-4)-(180)=0

We multiply parentheses ..

x^2+(+x^2-4x+3x-12)-3x-180=0

We get rid of parentheses

x^2+x^2-4x+3x-3x-12-180=0

We add all the numbers together, and all the variables

2x^2-4x-192=0

a = 2; b = -4; c = -192;
Δ = b2-4ac
Δ = -42-4·2·(-192)
Δ = 1552
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:

$\sqrt{\Delta}=\sqrt{1552}=\sqrt{16*97}=\sqrt{16}*\sqrt{97}=4\sqrt{97}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{97}}{2*2}=\frac{4-4\sqrt{97}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{97}}{2*2}=\frac{4+4\sqrt{97}}{4}
$