3z^2=127

Solution for 3z^2=127 equation:

3z^2=127

We move all terms to the left:

3z^2-(127)=0

a = 3; b = 0; c = -127;
Δ = b2-4ac
Δ = 02-4·3·(-127)
Δ = 1524
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:

$\sqrt{\Delta}=\sqrt{1524}=\sqrt{4*381}=\sqrt{4}*\sqrt{381}=2\sqrt{381}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{381}}{2*3}=\frac{0-2\sqrt{381}}{6}
=-\frac{2\sqrt{381}}{6}
=-\frac{\sqrt{381}}{3}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{381}}{2*3}=\frac{0+2\sqrt{381}}{6}
=\frac{2\sqrt{381}}{6}
=\frac{\sqrt{381}}{3}
$