3z^2+9z-3=0

Solution for 3z^2+9z-3=0 equation:

3z^2+9z-3=0

a = 3; b = 9; c = -3;
Δ = b2-4ac
Δ = 92-4·3·(-3)
Δ = 117
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:

$\sqrt{\Delta}=\sqrt{117}=\sqrt{9*13}=\sqrt{9}*\sqrt{13}=3\sqrt{13}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-3\sqrt{13}}{2*3}=\frac{-9-3\sqrt{13}}{6}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+3\sqrt{13}}{2*3}=\frac{-9+3\sqrt{13}}{6}
$