-v^2-16v+132=-10v-3

Solution for -v^2-16v+132=-10v-3 equation:

-v^2-16v+132=-10v-3

We move all terms to the left:

-v^2-16v+132-(-10v-3)=0

We add all the numbers together, and all the variables

-1v^2-16v-(-10v-3)+132=0

We get rid of parentheses

-1v^2-16v+10v+3+132=0

We add all the numbers together, and all the variables

-1v^2-6v+135=0

a = -1; b = -6; c = +135;
Δ = b2-4ac
Δ = -62-4·(-1)·135
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{-(-6)-24}{2*-1}=\frac{-18}{-2}
=+9
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+24}{2*-1}=\frac{30}{-2}
=-15
$