(V+3)^2=2v^2+8v+1

Solution for (V+3)^2=2v^2+8v+1 equation:

(+3)^2=2V^2+8V+1

We move all terms to the left:

(+3)^2-(2V^2+8V+1)=0

We add all the numbers together, and all the variables

-(2V^2+8V+1)+3^2=0

We add all the numbers together, and all the variables

-(2V^2+8V+1)+9=0

We get rid of parentheses

-2V^2-8V-1+9=0

We add all the numbers together, and all the variables

-2V^2-8V+8=0

a = -2; b = -8; c = +8;
Δ = b2-4ac
Δ = -82-4·(-2)·8
Δ = 128
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:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$V_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8\sqrt{2}}{2*-2}=\frac{8-8\sqrt{2}}{-4}
$
$V_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8\sqrt{2}}{2*-2}=\frac{8+8\sqrt{2}}{-4}
$