n^2-99=0

Solution for n^2-99=0 equation:

n^2-99=0

a = 1; b = 0; c = -99;
Δ = b2-4ac
Δ = 02-4·1·(-99)
Δ = 396
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:

$\sqrt{\Delta}=\sqrt{396}=\sqrt{36*11}=\sqrt{36}*\sqrt{11}=6\sqrt{11}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{11}}{2*1}=\frac{0-6\sqrt{11}}{2}
=-\frac{6\sqrt{11}}{2}
=-3\sqrt{11}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{11}}{2*1}=\frac{0+6\sqrt{11}}{2}
=\frac{6\sqrt{11}}{2}
=3\sqrt{11}
$