45=1.3v+.06v^2

Solution for 45=1.3v+.06v^2 equation:

45=1.3v+.06v^2

We move all terms to the left:

45-(1.3v+.06v^2)=0

We get rid of parentheses

-.06v^2-1.3v+45=0

We add all the numbers together, and all the variables

-0.06v^2-1.3v+45=0

a = -0.06; b = -1.3; c = +45;
Δ = b2-4ac
Δ = -1.32-4·(-0.06)·45
Δ = 12.49
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{12.49}}{2*-0.06}=\frac{1.3-\sqrt{12.49}}{-0.12}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1.3)+\sqrt{12.49}}{2*-0.06}=\frac{1.3+\sqrt{12.49}}{-0.12}
$