V=4v^2+14v+12

Solution for V=4v^2+14v+12 equation:

=4V^2+14V+12

We move all terms to the left:

-(4V^2+14V+12)=0

We get rid of parentheses

-4V^2-14V-12=0

a = -4; b = -14; c = -12;
Δ = b2-4ac
Δ = -142-4·(-4)·(-12)
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$V_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$V_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$V_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2}{2*-4}=\frac{12}{-8}
=-1+1/2
$
$V_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2}{2*-4}=\frac{16}{-8}
=-2
$