400=0.05v^2+2.2v

Solution for 400=0.05v^2+2.2v equation:

400=0.05v^2+2.2v

We move all terms to the left:

400-(0.05v^2+2.2v)=0

We get rid of parentheses

-0.05v^2-2.2v+400=0

a = -0.05; b = -2.2; c = +400;
Δ = b2-4ac
Δ = -2.22-4·(-0.05)·400
Δ = 84.84
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{-(-2.2)-\sqrt{84.84}}{2*-0.05}=\frac{2.2-\sqrt{84.84}}{-0.1}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2.2)+\sqrt{84.84}}{2*-0.05}=\frac{2.2+\sqrt{84.84}}{-0.1}
$