x^2+18x=99

Solution for x^2+18x=99 equation:

x^2+18x=99

We move all terms to the left:

x^2+18x-(99)=0

a = 1; b = 18; c = -99;
Δ = b2-4ac
Δ = 182-4·1·(-99)
Δ = 720
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:

$\sqrt{\Delta}=\sqrt{720}=\sqrt{144*5}=\sqrt{144}*\sqrt{5}=12\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-12\sqrt{5}}{2*1}=\frac{-18-12\sqrt{5}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+12\sqrt{5}}{2*1}=\frac{-18+12\sqrt{5}}{2}
$