-0.1x^2+1.1x+0.5=0

Solution for -0.1x^2+1.1x+0.5=0 equation:

-0.1x^2+1.1x+0.5=0

a = -0.1; b = 1.1; c = +0.5;
Δ = b2-4ac
Δ = 1.12-4·(-0.1)·0.5
Δ = 1.41
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{-(1.1)-\sqrt{1.41}}{2*-0.1}=\frac{-1.1-\sqrt{1.41}}{-0.2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1.1)+\sqrt{1.41}}{2*-0.1}=\frac{-1.1+\sqrt{1.41}}{-0.2}
$