2v^2+8v+5=0

Solution for 2v^2+8v+5=0 equation:

2v^2+8v+5=0

a = 2; b = 8; c = +5;
Δ = b2-4ac
Δ = 82-4·2·5
Δ = 24
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:

$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{6}}{2*2}=\frac{-8-2\sqrt{6}}{4}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{6}}{2*2}=\frac{-8+2\sqrt{6}}{4}
$