5y2-28=0

Solution for 5y2-28=0 equation:

5y^2-28=0

a = 5; b = 0; c = -28;
Δ = b2-4ac
Δ = 02-4·5·(-28)
Δ = 560
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:

$\sqrt{\Delta}=\sqrt{560}=\sqrt{16*35}=\sqrt{16}*\sqrt{35}=4\sqrt{35}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{35}}{2*5}=\frac{0-4\sqrt{35}}{10}
=-\frac{4\sqrt{35}}{10}
=-\frac{2\sqrt{35}}{5}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{35}}{2*5}=\frac{0+4\sqrt{35}}{10}
=\frac{4\sqrt{35}}{10}
=\frac{2\sqrt{35}}{5}
$