4q-q^2+2=0

Solution for 4q-q^2+2=0 equation:

4q-q^2+2=0

We add all the numbers together, and all the variables

-1q^2+4q+2=0

a = -1; b = 4; c = +2;
Δ = b2-4ac
Δ = 42-4·(-1)·2
Δ = 24
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{6}}{2*-1}=\frac{-4-2\sqrt{6}}{-2}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{6}}{2*-1}=\frac{-4+2\sqrt{6}}{-2}
$