3x^2-242=0

Solution for 3x^2-242=0 equation:

3x^2-242=0

a = 3; b = 0; c = -242;
Δ = b2-4ac
Δ = 02-4·3·(-242)
Δ = 2904
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:

$\sqrt{\Delta}=\sqrt{2904}=\sqrt{484*6}=\sqrt{484}*\sqrt{6}=22\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-22\sqrt{6}}{2*3}=\frac{0-22\sqrt{6}}{6}
=-\frac{22\sqrt{6}}{6}
=-\frac{11\sqrt{6}}{3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+22\sqrt{6}}{2*3}=\frac{0+22\sqrt{6}}{6}
=\frac{22\sqrt{6}}{6}
=\frac{11\sqrt{6}}{3}
$