-w^2+4w+5=0

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

-w^2+4w+5=0

We add all the numbers together, and all the variables

-1w^2+4w+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:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-6}{2*-1}=\frac{-10}{-2}
=+5
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+6}{2*-1}=\frac{2}{-2}
=-1
$