V^2=2v^2+8v+5

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

^2=2V^2+8V+5

We move all terms to the left:

^2-(2V^2+8V+5)=0

We add all the numbers together, and all the variables

-(2V^2+8V+5)=0

We get rid of parentheses

-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}
$