-0.002x^2+0.5x+10=0

Solution for -0.002x^2+0.5x+10=0 equation:

-0.002x^2+0.5x+10=0

a = -0.002; b = 0.5; c = +10;
Δ = b2-4ac
Δ = 0.52-4·(-0.002)·10
Δ = 0.33
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0.5)-\sqrt{0.33}}{2*-0.002}=\frac{-0.5-\sqrt{0.33}}{-0.004}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.5)+\sqrt{0.33}}{2*-0.002}=\frac{-0.5+\sqrt{0.33}}{-0.004}
$