250=0.05v^2+1.1v

Solution for 250=0.05v^2+1.1v equation:

250=0.05v^2+1.1v

We move all terms to the left:

250-(0.05v^2+1.1v)=0

We get rid of parentheses

-0.05v^2-1.1v+250=0

a = -0.05; b = -1.1; c = +250;
Δ = b2-4ac
Δ = -1.12-4·(-0.05)·250
Δ = 51.21
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{51.21}}{2*-0.05}=\frac{1.1-\sqrt{51.21}}{-0.1}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1.1)+\sqrt{51.21}}{2*-0.05}=\frac{1.1+\sqrt{51.21}}{-0.1}
$