s(s+12)=21

Solution for s(s+12)=21 equation:

s(s+12)=21

We move all terms to the left:

s(s+12)-(21)=0

We multiply parentheses

s^2+12s-21=0

a = 1; b = 12; c = -21;
Δ = b2-4ac
Δ = 122-4·1·(-21)
Δ = 228
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:

$\sqrt{\Delta}=\sqrt{228}=\sqrt{4*57}=\sqrt{4}*\sqrt{57}=2\sqrt{57}$
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{57}}{2*1}=\frac{-12-2\sqrt{57}}{2}
$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{57}}{2*1}=\frac{-12+2\sqrt{57}}{2}
$