10n^2+2=290

Solution for 10n^2+2=290 equation:

10n^2+2=290

We move all terms to the left:

10n^2+2-(290)=0

We add all the numbers together, and all the variables

10n^2-288=0

a = 10; b = 0; c = -288;
Δ = b2-4ac
Δ = 02-4·10·(-288)
Δ = 11520
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:

$\sqrt{\Delta}=\sqrt{11520}=\sqrt{2304*5}=\sqrt{2304}*\sqrt{5}=48\sqrt{5}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-48\sqrt{5}}{2*10}=\frac{0-48\sqrt{5}}{20}
=-\frac{48\sqrt{5}}{20}
=-\frac{12\sqrt{5}}{5}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+48\sqrt{5}}{2*10}=\frac{0+48\sqrt{5}}{20}
=\frac{48\sqrt{5}}{20}
=\frac{12\sqrt{5}}{5}
$