8v^2+16v-28=0

Solution for 8v^2+16v-28=0 equation:

8v^2+16v-28=0

a = 8; b = 16; c = -28;
Δ = b2-4ac
Δ = 162-4·8·(-28)
Δ = 1152
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:

$\sqrt{\Delta}=\sqrt{1152}=\sqrt{576*2}=\sqrt{576}*\sqrt{2}=24\sqrt{2}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-24\sqrt{2}}{2*8}=\frac{-16-24\sqrt{2}}{16}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+24\sqrt{2}}{2*8}=\frac{-16+24\sqrt{2}}{16}
$