2q-q^2-0.01=0

Solution for 2q-q^2-0.01=0 equation:

2q-q^2-0.01=0

We add all the numbers together, and all the variables

-1q^2+2q-0.01=0

a = -1; b = 2; c = -0.01;
Δ = b2-4ac
Δ = 22-4·(-1)·(-0.01)
Δ = 3.96
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-\sqrt{3.96}}{2*-1}=\frac{-2-\sqrt{3.96}}{-2}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+\sqrt{3.96}}{2*-1}=\frac{-2+\sqrt{3.96}}{-2}
$