x^2+18x=98

Solution for x^2+18x=98 equation:

x^2+18x=98

We move all terms to the left:

x^2+18x-(98)=0

a = 1; b = 18; c = -98;
Δ = b2-4ac
Δ = 182-4·1·(-98)
Δ = 716
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:

$\sqrt{\Delta}=\sqrt{716}=\sqrt{4*179}=\sqrt{4}*\sqrt{179}=2\sqrt{179}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{179}}{2*1}=\frac{-18-2\sqrt{179}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{179}}{2*1}=\frac{-18+2\sqrt{179}}{2}
$