28=8v^2

Solution for 28=8v^2 equation:

28=8v^2

We move all terms to the left:

28-(8v^2)=0

a = -8; b = 0; c = +28;
Δ = b2-4ac
Δ = 02-4·(-8)·28
Δ = 896
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{896}=\sqrt{64*14}=\sqrt{64}*\sqrt{14}=8\sqrt{14}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{14}}{2*-8}=\frac{0-8\sqrt{14}}{-16}
=-\frac{8\sqrt{14}}{-16}
=-\frac{\sqrt{14}}{-2}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{14}}{2*-8}=\frac{0+8\sqrt{14}}{-16}
=\frac{8\sqrt{14}}{-16}
=\frac{\sqrt{14}}{-2}
$