3n2=17

Solution for 3n2=17 equation:

3n^2=17

We move all terms to the left:

3n^2-(17)=0

a = 3; b = 0; c = -17;
Δ = b2-4ac
Δ = 02-4·3·(-17)
Δ = 204
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:

$\sqrt{\Delta}=\sqrt{204}=\sqrt{4*51}=\sqrt{4}*\sqrt{51}=2\sqrt{51}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{51}}{2*3}=\frac{0-2\sqrt{51}}{6}
=-\frac{2\sqrt{51}}{6}
=-\frac{\sqrt{51}}{3}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{51}}{2*3}=\frac{0+2\sqrt{51}}{6}
=\frac{2\sqrt{51}}{6}
=\frac{\sqrt{51}}{3}
$