-v^2+128=10v+9

Solution for -v^2+128=10v+9 equation:

-v^2+128=10v+9

We move all terms to the left:

-v^2+128-(10v+9)=0

We add all the numbers together, and all the variables

-1v^2-(10v+9)+128=0

We get rid of parentheses

-1v^2-10v-9+128=0

We add all the numbers together, and all the variables

-1v^2-10v+119=0

a = -1; b = -10; c = +119;
Δ = b2-4ac
Δ = -102-4·(-1)·119
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-24}{2*-1}=\frac{-14}{-2}
=+7
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+24}{2*-1}=\frac{34}{-2}
=-17
$