-1.125x2+2.5x=0

Solution for -1.125x2+2.5x=0 equation:

-1.125x^2+2.5x=0

a = -1.125; b = 2.5; c = 0;
Δ = b2-4ac
Δ = 2.52-4·(-1.125)·0
Δ = 6.25
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{-(2.5)-\sqrt{6.25}}{2*-1.125}=\frac{-2.5-\sqrt{6.25}}{-2.25}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2.5)+\sqrt{6.25}}{2*-1.125}=\frac{-2.5+\sqrt{6.25}}{-2.25}
$