2n^2+4=202

Solution for 2n^2+4=202 equation:

2n^2+4=202

We move all terms to the left:

2n^2+4-(202)=0

We add all the numbers together, and all the variables

2n^2-198=0

a = 2; b = 0; c = -198;
Δ = b2-4ac
Δ = 02-4·2·(-198)
Δ = 1584
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:

$\sqrt{\Delta}=\sqrt{1584}=\sqrt{144*11}=\sqrt{144}*\sqrt{11}=12\sqrt{11}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{11}}{2*2}=\frac{0-12\sqrt{11}}{4}
=-\frac{12\sqrt{11}}{4}
=-3\sqrt{11}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{11}}{2*2}=\frac{0+12\sqrt{11}}{4}
=\frac{12\sqrt{11}}{4}
=3\sqrt{11}
$