-0.01x^2+x=-3

Solution for -0.01x^2+x=-3 equation:

-0.01x^2+x=-3

We move all terms to the left:

-0.01x^2+x-(-3)=0

We add all the numbers together, and all the variables

-0.01x^2+x+3=0

a = -0.01; b = 1; c = +3;
Δ = b2-4ac
Δ = 12-4·(-0.01)·3
Δ = 1.12
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{1.12}}{2*-0.01}=\frac{-1-\sqrt{1.12}}{-0.02}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{1.12}}{2*-0.01}=\frac{-1+\sqrt{1.12}}{-0.02}
$