2n^2+4=22

Solution for 2n^2+4=22 equation:

2n^2+4=22

We move all terms to the left:

2n^2+4-(22)=0

We add all the numbers together, and all the variables

2n^2-18=0

a = 2; b = 0; c = -18;
Δ = b2-4ac
Δ = 02-4·2·(-18)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12}{2*2}=\frac{-12}{4}
=-3
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12}{2*2}=\frac{12}{4}
=3
$