12=-6+0.5v^2

Solution for 12=-6+0.5v^2 equation:

12=-6+0.5v^2

We move all terms to the left:

12-(-6+0.5v^2)=0

We get rid of parentheses

-0.5v^2+6+12=0

We add all the numbers together, and all the variables

-0.5v^2+18=0

a = -0.5; b = 0; c = +18;
Δ = b2-4ac
Δ = 02-4·(-0.5)·18
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6}{2*-0.5}=\frac{-6}{-1}
=+6
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6}{2*-0.5}=\frac{6}{-1}
=-6
$