11=1.3v+0.04v^2

Solution for 11=1.3v+0.04v^2 equation:

11=1.3v+0.04v^2

We move all terms to the left:

11-(1.3v+0.04v^2)=0

We get rid of parentheses

-0.04v^2-1.3v+11=0

a = -0.04; b = -1.3; c = +11;
Δ = b2-4ac
Δ = -1.32-4·(-0.04)·11
Δ = 3.45
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{-(-1.3)-\sqrt{3.45}}{2*-0.04}=\frac{1.3-\sqrt{3.45}}{-0.08}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1.3)+\sqrt{3.45}}{2*-0.04}=\frac{1.3+\sqrt{3.45}}{-0.08}
$