2+1.1x-0.1x^2=3

Solution for 2+1.1x-0.1x^2=3 equation:

2+1.1x-0.1x^2=3

We move all terms to the left:

2+1.1x-0.1x^2-(3)=0

We add all the numbers together, and all the variables

-0.1x^2+1.1x-1=0

a = -0.1; b = 1.1; c = -1;
Δ = b2-4ac
Δ = 1.12-4·(-0.1)·(-1)
Δ = 0.81
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{0.81}}{2*-0.1}=\frac{-1.1-\sqrt{0.81}}{-0.2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1.1)+\sqrt{0.81}}{2*-0.1}=\frac{-1.1+\sqrt{0.81}}{-0.2}
$