5y^2=21

Solution for 5y^2=21 equation:

5y^2=21

We move all terms to the left:

5y^2-(21)=0

a = 5; b = 0; c = -21;
Δ = b2-4ac
Δ = 02-4·5·(-21)
Δ = 420
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:

$\sqrt{\Delta}=\sqrt{420}=\sqrt{4*105}=\sqrt{4}*\sqrt{105}=2\sqrt{105}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{105}}{2*5}=\frac{0-2\sqrt{105}}{10}
=-\frac{2\sqrt{105}}{10}
=-\frac{\sqrt{105}}{5}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{105}}{2*5}=\frac{0+2\sqrt{105}}{10}
=\frac{2\sqrt{105}}{10}
=\frac{\sqrt{105}}{5}
$