-2.6x^2+1.6x=0

Solution for -2.6x^2+1.6x=0 equation:

-2.6x^2+1.6x=0

a = -2.6; b = 1.6; c = 0;
Δ = b2-4ac
Δ = 1.62-4·(-2.6)·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*-2.6}=\frac{-1.6-\sqrt{2.56}}{-5.2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1.6)+\sqrt{2.56}}{2*-2.6}=\frac{-1.6+\sqrt{2.56}}{-5.2}
$