3y^2=102

Solution for 3y^2=102 equation:

3y^2=102

We move all terms to the left:

3y^2-(102)=0

a = 3; b = 0; c = -102;
Δ = b2-4ac
Δ = 02-4·3·(-102)
Δ = 1224
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1224}=\sqrt{36*34}=\sqrt{36}*\sqrt{34}=6\sqrt{34}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{34}}{2*3}=\frac{0-6\sqrt{34}}{6}
=-\frac{6\sqrt{34}}{6}
=-\sqrt{34}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{34}}{2*3}=\frac{0+6\sqrt{34}}{6}
=\frac{6\sqrt{34}}{6}
=\sqrt{34}
$