0.002x^2+.6x+20=0

Solution for 0.002x^2+.6x+20=0 equation:

0.002x^2+.6x+20=0

a = 0.002; b = .6; c = +20;
Δ = b2-4ac
Δ = .62-4·0.002·20
Δ = 0.2
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{-(.6)-\sqrt{0.2}}{2*0.002}=\frac{-0.6-\sqrt{0.2}}{0.004}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(.6)+\sqrt{0.2}}{2*0.002}=\frac{-0.6+\sqrt{0.2}}{0.004}
$