0.1x^2+x-0.9=0

Solution for 0.1x^2+x-0.9=0 equation:

0.1x^2+x-0.9=0

a = 0.1; b = 1; c = -0.9;
Δ = b2-4ac
Δ = 12-4·0.1·(-0.9)
Δ = 1.36
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{1.36}}{2*0.1}=\frac{-1-\sqrt{1.36}}{0.2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{1.36}}{2*0.1}=\frac{-1+\sqrt{1.36}}{0.2}
$