n^2=12n+21

Solution for n^2=12n+21 equation:

n^2=12n+21

We move all terms to the left:

n^2-(12n+21)=0

We get rid of parentheses

n^2-12n-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:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

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