142=3z^2

Solution for 142=3z^2 equation:

142=3z^2

We move all terms to the left:

142-(3z^2)=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}
$