4v^2+8v=12

Solution for 4v^2+8v=12 equation:

4v^2+8v=12

We move all terms to the left:

4v^2+8v-(12)=0

a = 4; b = 8; c = -12;
Δ = b2-4ac
Δ = 82-4·4·(-12)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-16}{2*4}=\frac{-24}{8}
=-3
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+16}{2*4}=\frac{8}{8}
=1
$