3x^2=328

Solution for 3x^2=328 equation:

3x^2=328

We move all terms to the left:

3x^2-(328)=0

a = 3; b = 0; c = -328;
Δ = b2-4ac
Δ = 02-4·3·(-328)
Δ = 3936
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{3936}=\sqrt{16*246}=\sqrt{16}*\sqrt{246}=4\sqrt{246}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{246}}{2*3}=\frac{0-4\sqrt{246}}{6}
=-\frac{4\sqrt{246}}{6}
=-\frac{2\sqrt{246}}{3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{246}}{2*3}=\frac{0+4\sqrt{246}}{6}
=\frac{4\sqrt{246}}{6}
=\frac{2\sqrt{246}}{3}
$