-0.02x^2+1.1x=3.12

Solution for -0.02x^2+1.1x=3.12 equation:

-0.02x^2+1.1x=3.12

We move all terms to the left:

-0.02x^2+1.1x-(3.12)=0

We add all the numbers together, and all the variables

-0.02x^2+1.1x-3.12=0

a = -0.02; b = 1.1; c = -3.12;
Δ = b2-4ac
Δ = 1.12-4·(-0.02)·(-3.12)
Δ = 0.9604
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{0.9604}}{2*-0.02}=\frac{-1.1-\sqrt{0.9604}}{-0.04}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1.1)+\sqrt{0.9604}}{2*-0.02}=\frac{-1.1+\sqrt{0.9604}}{-0.04}
$