w^2+0.5w-216=0

Solution for w^2+0.5w-216=0 equation:

w^2+0.5w-216=0

a = 1; b = 0.5; c = -216;
Δ = b2-4ac
Δ = 0.52-4·1·(-216)
Δ = 864.25
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{-(0.5)-\sqrt{864.25}}{2*1}=\frac{-0.5-\sqrt{864.25}}{2}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.5)+\sqrt{864.25}}{2*1}=\frac{-0.5+\sqrt{864.25}}{2}
$