s^2-18s=40

Solution for s^2-18s=40 equation:

s^2-18s=40

We move all terms to the left:

s^2-18s-(40)=0

a = 1; b = -18; c = -40;
Δ = b2-4ac
Δ = -182-4·1·(-40)
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-22}{2*1}=\frac{-4}{2}
=-2
$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+22}{2*1}=\frac{40}{2}
=20
$