X^2-3x+(x+4)(x-5)=180

Solution for X^2-3x+(x+4)(x-5)=180 equation:

X^2-3X+(X+4)(X-5)=180

We move all terms to the left:

X^2-3X+(X+4)(X-5)-(180)=0

We multiply parentheses ..

X^2+(+X^2-5X+4X-20)-3X-180=0

We get rid of parentheses

X^2+X^2-5X+4X-3X-20-180=0

We add all the numbers together, and all the variables

2X^2-4X-200=0

a = 2; b = -4; c = -200;
Δ = b2-4ac
Δ = -42-4·2·(-200)
Δ = 1616
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:

$\sqrt{\Delta}=\sqrt{1616}=\sqrt{16*101}=\sqrt{16}*\sqrt{101}=4\sqrt{101}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{101}}{2*2}=\frac{4-4\sqrt{101}}{4}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{101}}{2*2}=\frac{4+4\sqrt{101}}{4}
$