3z^2=4

Solution for 3z^2=4 equation:

3z^2=4

We move all terms to the left:

3z^2-(4)=0

a = 3; b = 0; c = -4;
Δ = b2-4ac
Δ = 02-4·3·(-4)
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{3}}{2*3}=\frac{0-4\sqrt{3}}{6}
=-\frac{4\sqrt{3}}{6}
=-\frac{2\sqrt{3}}{3}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{3}}{2*3}=\frac{0+4\sqrt{3}}{6}
=\frac{4\sqrt{3}}{6}
=\frac{2\sqrt{3}}{3}
$