-v^2+14v-40=0

Solution for -v^2+14v-40=0 equation:

-v^2+14v-40=0

We add all the numbers together, and all the variables

-1v^2+14v-40=0

a = -1; b = 14; c = -40;
Δ = b2-4ac
Δ = 142-4·(-1)·(-40)
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-6}{2*-1}=\frac{-20}{-2}
=+10
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+6}{2*-1}=\frac{-8}{-2}
=+4
$