3x^2-490=0

Solution for 3x^2-490=0 equation:

3x^2-490=0

a = 3; b = 0; c = -490;
Δ = b2-4ac
Δ = 02-4·3·(-490)
Δ = 5880
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:

$\sqrt{\Delta}=\sqrt{5880}=\sqrt{196*30}=\sqrt{196}*\sqrt{30}=14\sqrt{30}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-14\sqrt{30}}{2*3}=\frac{0-14\sqrt{30}}{6}
=-\frac{14\sqrt{30}}{6}
=-\frac{7\sqrt{30}}{3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+14\sqrt{30}}{2*3}=\frac{0+14\sqrt{30}}{6}
=\frac{14\sqrt{30}}{6}
=\frac{7\sqrt{30}}{3}
$