12=0.002(2v^2+10v+3000)

Solution for 12=0.002(2v^2+10v+3000) equation:

12=0.002(2v^2+10v+3000)

We move all terms to the left:

12-(0.002(2v^2+10v+3000))=0


We calculate terms in parentheses: -(0.002(2v^2+10v+3000)), so:

0.002(2v^2+10v+3000)

We multiply parentheses

0.004v^2+0.02v+6

Back to the equation:

-(0.004v^2+0.02v+6)


We get rid of parentheses

-0.004v^2-0.02v-6+12=0

We add all the numbers together, and all the variables

-0.004v^2-0.02v+6=0

a = -0.004; b = -0.02; c = +6;
Δ = b2-4ac
Δ = -0.022-4·(-0.004)·6
Δ = 0.0964
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{-(-0.02)-\sqrt{0.0964}}{2*-0.004}=\frac{0.02-\sqrt{0.0964}}{-0.008}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-0.02)+\sqrt{0.0964}}{2*-0.004}=\frac{0.02+\sqrt{0.0964}}{-0.008}
$