0=x^2+18x-120

Solution for 0=x^2+18x-120 equation:

0=x^2+18x-120

We move all terms to the left:

0-(x^2+18x-120)=0

We add all the numbers together, and all the variables

-(x^2+18x-120)=0

We get rid of parentheses

-x^2-18x+120=0

We add all the numbers together, and all the variables

-1x^2-18x+120=0

a = -1; b = -18; c = +120;
Δ = b2-4ac
Δ = -182-4·(-1)·120
Δ = 804
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:

$\sqrt{\Delta}=\sqrt{804}=\sqrt{4*201}=\sqrt{4}*\sqrt{201}=2\sqrt{201}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{201}}{2*-1}=\frac{18-2\sqrt{201}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{201}}{2*-1}=\frac{18+2\sqrt{201}}{-2}
$