-w^2+3w+50=0

Solution for -w^2+3w+50=0 equation:

-w^2+3w+50=0

We add all the numbers together, and all the variables

-1w^2+3w+50=0

a = -1; b = 3; c = +50;
Δ = b2-4ac
Δ = 32-4·(-1)·50
Δ = 209
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{209}}{2*-1}=\frac{-3-\sqrt{209}}{-2}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{209}}{2*-1}=\frac{-3+\sqrt{209}}{-2}
$