300=0.02v^2+1.2v

Solution for 300=0.02v^2+1.2v equation:

300=0.02v^2+1.2v

We move all terms to the left:

300-(0.02v^2+1.2v)=0

We get rid of parentheses

-0.02v^2-1.2v+300=0

a = -0.02; b = -1.2; c = +300;
Δ = b2-4ac
Δ = -1.22-4·(-0.02)·300
Δ = 25.44
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{-(-1.2)-\sqrt{25.44}}{2*-0.02}=\frac{1.2-\sqrt{25.44}}{-0.04}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1.2)+\sqrt{25.44}}{2*-0.02}=\frac{1.2+\sqrt{25.44}}{-0.04}
$