x^2+18x=228=0

Solution for x^2+18x=228=0 equation:

x^2+18x=228=0

We move all terms to the left:

x^2+18x-(228)=0

a = 1; b = 18; c = -228;
Δ = b2-4ac
Δ = 182-4·1·(-228)
Δ = 1236
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:

$\sqrt{\Delta}=\sqrt{1236}=\sqrt{4*309}=\sqrt{4}*\sqrt{309}=2\sqrt{309}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{309}}{2*1}=\frac{-18-2\sqrt{309}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{309}}{2*1}=\frac{-18+2\sqrt{309}}{2}
$