-0.06x^2+1.6x=0

Solution for -0.06x^2+1.6x=0 equation:

-0.06x^2+1.6x=0

a = -0.06; b = 1.6; c = 0;
Δ = b2-4ac
Δ = 1.62-4·(-0.06)·0
Δ = 2.56
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{-(1.6)-\sqrt{2.56}}{2*-0.06}=\frac{-1.6-\sqrt{2.56}}{-0.12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1.6)+\sqrt{2.56}}{2*-0.06}=\frac{-1.6+\sqrt{2.56}}{-0.12}
$