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

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

X^2-2X+(X+2)(X-3)=180

We move all terms to the left:

X^2-2X+(X+2)(X-3)-(180)=0

We multiply parentheses ..

X^2+(+X^2-3X+2X-6)-2X-180=0

We get rid of parentheses

X^2+X^2-3X+2X-2X-6-180=0

We add all the numbers together, and all the variables

2X^2-3X-186=0

a = 2; b = -3; c = -186;
Δ = b2-4ac
Δ = -32-4·2·(-186)
Δ = 1497
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{1497}}{2*2}=\frac{3-\sqrt{1497}}{4}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{1497}}{2*2}=\frac{3+\sqrt{1497}}{4}
$