3z^2-4=140

Solution for 3z^2-4=140 equation:

3z^2-4=140

We move all terms to the left:

3z^2-4-(140)=0

We add all the numbers together, and all the variables

3z^2-144=0

a = 3; b = 0; c = -144;
Δ = b2-4ac
Δ = 02-4·3·(-144)
Δ = 1728
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:

$\sqrt{\Delta}=\sqrt{1728}=\sqrt{576*3}=\sqrt{576}*\sqrt{3}=24\sqrt{3}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-24\sqrt{3}}{2*3}=\frac{0-24\sqrt{3}}{6}
=-\frac{24\sqrt{3}}{6}
=-4\sqrt{3}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+24\sqrt{3}}{2*3}=\frac{0+24\sqrt{3}}{6}
=\frac{24\sqrt{3}}{6}
=4\sqrt{3}
$