-0.5x^2+50=0

Solution for -0.5x^2+50=0 equation:

-0.5x^2+50=0

a = -0.5; b = 0; c = +50;
Δ = b2-4ac
Δ = 02-4·(-0.5)·50
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10}{2*-0.5}=\frac{-10}{-1}
=+10
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10}{2*-0.5}=\frac{10}{-1}
=-10
$