-v^2+2v^2+13v=0

Solution for -v^2+2v^2+13v=0 equation:

-v^2+2v^2+13v=0

We add all the numbers together, and all the variables

v^2+13v=0

a = 1; b = 13; c = 0;
Δ = b2-4ac
Δ = 132-4·1·0
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-13}{2*1}=\frac{-26}{2}
=-13
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+13}{2*1}=\frac{0}{2}
=0
$