-v^2-4v+5=0

Solution for -v^2-4v+5=0 equation:

-v^2-4v+5=0

We add all the numbers together, and all the variables

-1v^2-4v+5=0

a = -1; b = -4; c = +5;
Δ = b2-4ac
Δ = -42-4·(-1)·5
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{-(-4)-6}{2*-1}=\frac{-2}{-2}
=1
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+6}{2*-1}=\frac{10}{-2}
=-5
$