(V+5)^2=2v^2+15v+19

Solution for (V+5)^2=2v^2+15v+19 equation:

(+5)^2=2V^2+15V+19

We move all terms to the left:

(+5)^2-(2V^2+15V+19)=0

We add all the numbers together, and all the variables

-(2V^2+15V+19)+5^2=0

We add all the numbers together, and all the variables

-(2V^2+15V+19)+25=0

We get rid of parentheses

-2V^2-15V-19+25=0

We add all the numbers together, and all the variables

-2V^2-15V+6=0

a = -2; b = -15; c = +6;
Δ = b2-4ac
Δ = -152-4·(-2)·6
Δ = 273
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$V_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$V_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$V_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-\sqrt{273}}{2*-2}=\frac{15-\sqrt{273}}{-4}
$
$V_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{273}}{2*-2}=\frac{15+\sqrt{273}}{-4}
$