-v^2+12v=0

Solution for -v^2+12v=0 equation:

-v^2+12v=0

We add all the numbers together, and all the variables

-1v^2+12v=0

a = -1; b = 12; c = 0;
Δ = b2-4ac
Δ = 122-4·(-1)·0
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{-(12)-12}{2*-1}=\frac{-24}{-2}
=+12
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12}{2*-1}=\frac{0}{-2}
=0
$