-0.004x^2+0.6x=0

Solution for -0.004x^2+0.6x=0 equation:

-0.004x^2+0.6x=0

a = -0.004; b = 0.6; c = 0;
Δ = b2-4ac
Δ = 0.62-4·(-0.004)·0
Δ = 0.36
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{-(0.6)-\sqrt{0.36}}{2*-0.004}=\frac{-0.6-\sqrt{0.36}}{-0.008}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.6)+\sqrt{0.36}}{2*-0.004}=\frac{-0.6+\sqrt{0.36}}{-0.008}
$