(V+1)^2=2v^2+5v+3

Solution for (V+1)^2=2v^2+5v+3 equation:

(+1)^2=2V^2+5V+3

We move all terms to the left:

(+1)^2-(2V^2+5V+3)=0

We add all the numbers together, and all the variables

-(2V^2+5V+3)+1^2=0

We add all the numbers together, and all the variables

-(2V^2+5V+3)+1=0

We get rid of parentheses

-2V^2-5V-3+1=0

We add all the numbers together, and all the variables

-2V^2-5V-2=0

a = -2; b = -5; c = -2;
Δ = b2-4ac
Δ = -52-4·(-2)·(-2)
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$V_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$V_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$V_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-3}{2*-2}=\frac{2}{-4}
=-1/2
$
$V_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+3}{2*-2}=\frac{8}{-4}
=-2
$