20n^2+2=292

Solution for 20n^2+2=292 equation:

20n^2+2=292

We move all terms to the left:

20n^2+2-(292)=0

We add all the numbers together, and all the variables

20n^2-290=0

a = 20; b = 0; c = -290;
Δ = b2-4ac
Δ = 02-4·20·(-290)
Δ = 23200
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:

$\sqrt{\Delta}=\sqrt{23200}=\sqrt{400*58}=\sqrt{400}*\sqrt{58}=20\sqrt{58}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20\sqrt{58}}{2*20}=\frac{0-20\sqrt{58}}{40}
=-\frac{20\sqrt{58}}{40}
=-\frac{\sqrt{58}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20\sqrt{58}}{2*20}=\frac{0+20\sqrt{58}}{40}
=\frac{20\sqrt{58}}{40}
=\frac{\sqrt{58}}{2}
$