s^2+16s=90

Solution for s^2+16s=90 equation:

s^2+16s=90

We move all terms to the left:

s^2+16s-(90)=0

a = 1; b = 16; c = -90;
Δ = b2-4ac
Δ = 162-4·1·(-90)
Δ = 616
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{616}=\sqrt{4*154}=\sqrt{4}*\sqrt{154}=2\sqrt{154}$
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{154}}{2*1}=\frac{-16-2\sqrt{154}}{2}
$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{154}}{2*1}=\frac{-16+2\sqrt{154}}{2}
$