4v^2-28v+28=0

Solution for 4v^2-28v+28=0 equation:

4v^2-28v+28=0

a = 4; b = -28; c = +28;
Δ = b2-4ac
Δ = -282-4·4·28
Δ = 336
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:

$\sqrt{\Delta}=\sqrt{336}=\sqrt{16*21}=\sqrt{16}*\sqrt{21}=4\sqrt{21}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-4\sqrt{21}}{2*4}=\frac{28-4\sqrt{21}}{8}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+4\sqrt{21}}{2*4}=\frac{28+4\sqrt{21}}{8}
$