2n^2=21

Solution for 2n^2=21 equation:

2n^2=21

We move all terms to the left:

2n^2-(21)=0

a = 2; b = 0; c = -21;
Δ = b2-4ac
Δ = 02-4·2·(-21)
Δ = 168
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:

$\sqrt{\Delta}=\sqrt{168}=\sqrt{4*42}=\sqrt{4}*\sqrt{42}=2\sqrt{42}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{42}}{2*2}=\frac{0-2\sqrt{42}}{4}
=-\frac{2\sqrt{42}}{4}
=-\frac{\sqrt{42}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{42}}{2*2}=\frac{0+2\sqrt{42}}{4}
=\frac{2\sqrt{42}}{4}
=\frac{\sqrt{42}}{2}
$