2n2=99

Solution for 2n2=99 equation:

2n^2=99

We move all terms to the left:

2n^2-(99)=0

a = 2; b = 0; c = -99;
Δ = b2-4ac
Δ = 02-4·2·(-99)
Δ = 792
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:

$\sqrt{\Delta}=\sqrt{792}=\sqrt{36*22}=\sqrt{36}*\sqrt{22}=6\sqrt{22}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{22}}{2*2}=\frac{0-6\sqrt{22}}{4}
=-\frac{6\sqrt{22}}{4}
=-\frac{3\sqrt{22}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{22}}{2*2}=\frac{0+6\sqrt{22}}{4}
=\frac{6\sqrt{22}}{4}
=\frac{3\sqrt{22}}{2}
$