142=3z^2-5

Solution for 142=3z^2-5 equation:

142=3z^2-5

We move all terms to the left:

142-(3z^2-5)=0

We get rid of parentheses

-3z^2+5+142=0

We add all the numbers together, and all the variables

-3z^2+147=0

a = -3; b = 0; c = +147;
Δ = b2-4ac
Δ = 02-4·(-3)·147
Δ = 1764
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1764}=42$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-42}{2*-3}=\frac{-42}{-6}
=+7
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+42}{2*-3}=\frac{42}{-6}
=-7
$