5y^2=27

Solution for 5y^2=27 equation:

5y^2=27

We move all terms to the left:

5y^2-(27)=0

a = 5; b = 0; c = -27;
Δ = b2-4ac
Δ = 02-4·5·(-27)
Δ = 540
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:

$\sqrt{\Delta}=\sqrt{540}=\sqrt{36*15}=\sqrt{36}*\sqrt{15}=6\sqrt{15}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{15}}{2*5}=\frac{0-6\sqrt{15}}{10}
=-\frac{6\sqrt{15}}{10}
=-\frac{3\sqrt{15}}{5}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{15}}{2*5}=\frac{0+6\sqrt{15}}{10}
=\frac{6\sqrt{15}}{10}
=\frac{3\sqrt{15}}{5}
$