-0.1x^2+x+5.5=0

Solution for -0.1x^2+x+5.5=0 equation:

-0.1x^2+x+5.5=0

a = -0.1; b = 1; c = +5.5;
Δ = b2-4ac
Δ = 12-4·(-0.1)·5.5
Δ = 3.2
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)-\sqrt{3.2}}{2*-0.1}=\frac{-1-\sqrt{3.2}}{-0.2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{3.2}}{2*-0.1}=\frac{-1+\sqrt{3.2}}{-0.2}
$