s^2+6s=27

Solution for s^2+6s=27 equation:

s^2+6s=27

We move all terms to the left:

s^2+6s-(27)=0

a = 1; b = 6; c = -27;
Δ = b2-4ac
Δ = 62-4·1·(-27)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-12}{2*1}=\frac{-18}{2}
=-9
$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+12}{2*1}=\frac{6}{2}
=3
$