2q-q^2=0.25

Solution for 2q-q^2=0.25 equation:

2q-q^2=0.25

We move all terms to the left:

2q-q^2-(0.25)=0

We add all the numbers together, and all the variables

-1q^2+2q-0.25=0

a = -1; b = 2; c = -0.25;
Δ = b2-4ac
Δ = 22-4·(-1)·(-0.25)
Δ = 3
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}}{2*-1}=\frac{-2-\sqrt{3}}{-2}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+\sqrt{3}}{2*-1}=\frac{-2+\sqrt{3}}{-2}
$