35=-2v+v^2

Solution for 35=-2v+v^2 equation:

35=-2v+v^2

We move all terms to the left:

35-(-2v+v^2)=0

We get rid of parentheses

-v^2+2v+35=0

We add all the numbers together, and all the variables

-1v^2+2v+35=0

a = -1; b = 2; c = +35;
Δ = b2-4ac
Δ = 22-4·(-1)·35
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-12}{2*-1}=\frac{-14}{-2}
=+7
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+12}{2*-1}=\frac{10}{-2}
=-5
$