3z2-142=0

Solution for 3z2-142=0 equation:

3z^2-142=0

a = 3; b = 0; c = -142;
Δ = b2-4ac
Δ = 02-4·3·(-142)
Δ = 1704
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1704}=\sqrt{4*426}=\sqrt{4}*\sqrt{426}=2\sqrt{426}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{426}}{2*3}=\frac{0-2\sqrt{426}}{6}
=-\frac{2\sqrt{426}}{6}
=-\frac{\sqrt{426}}{3}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{426}}{2*3}=\frac{0+2\sqrt{426}}{6}
=\frac{2\sqrt{426}}{6}
=\frac{\sqrt{426}}{3}
$