3z2-33=0

Solution for 3z2-33=0 equation:

3z^2-33=0

a = 3; b = 0; c = -33;
Δ = b2-4ac
Δ = 02-4·3·(-33)
Δ = 396
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:

$\sqrt{\Delta}=\sqrt{396}=\sqrt{36*11}=\sqrt{36}*\sqrt{11}=6\sqrt{11}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{11}}{2*3}=\frac{0-6\sqrt{11}}{6}
=-\frac{6\sqrt{11}}{6}
=-\sqrt{11}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{11}}{2*3}=\frac{0+6\sqrt{11}}{6}
=\frac{6\sqrt{11}}{6}
=\sqrt{11}
$