(V+5)=2v^2+14v+28

Solution for (V+5)=2v^2+14v+28 equation:

(+5)=2V^2+14V+28

We move all terms to the left:

(+5)-(2V^2+14V+28)=0

We add all the numbers together, and all the variables

-(2V^2+14V+28)+5=0

We get rid of parentheses

-2V^2-14V-28+5=0

We add all the numbers together, and all the variables

-2V^2-14V-23=0

a = -2; b = -14; c = -23;
Δ = b2-4ac
Δ = -142-4·(-2)·(-23)
Δ = 12
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:

$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$
$V_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2\sqrt{3}}{2*-2}=\frac{14-2\sqrt{3}}{-4}
$
$V_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2\sqrt{3}}{2*-2}=\frac{14+2\sqrt{3}}{-4}
$