-0.0006x^2+0.02x+20=0

Solution for -0.0006x^2+0.02x+20=0 equation:

-0.0006x^2+0.02x+20=0

a = -0.0006; b = 0.02; c = +20;
Δ = b2-4ac
Δ = 0.022-4·(-0.0006)·20
Δ = 0.0484
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{-(0.02)-\sqrt{0.0484}}{2*-0.0006}=\frac{-0.02-\sqrt{0.0484}}{-0.0012}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.02)+\sqrt{0.0484}}{2*-0.0006}=\frac{-0.02+\sqrt{0.0484}}{-0.0012}
$