.06v^2+1.1v=250

Solution for .06v^2+1.1v=250 equation:

.06v^2+1.1v=250

We move all terms to the left:

.06v^2+1.1v-(250)=0

a = .06; b = 1.1; c = -250;
Δ = b2-4ac
Δ = 1.12-4·.06·(-250)
Δ = 61.21
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{-(1.1)-\sqrt{61.21}}{2*.06}=\frac{-1.1-\sqrt{61.21}}{0.12}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1.1)+\sqrt{61.21}}{2*.06}=\frac{-1.1+\sqrt{61.21}}{0.12}
$