-2(q-4)=q^2

Solution for -2(q-4)=q^2 equation:

-2(q-4)=q^2

We move all terms to the left:

-2(q-4)-(q^2)=0

determiningTheFunctionDomain
-q^2-2(q-4)=0

We add all the numbers together, and all the variables

-1q^2-2(q-4)=0

We multiply parentheses

-1q^2-2q+8=0

a = -1; b = -2; c = +8;
Δ = b2-4ac
Δ = -22-4·(-1)·8
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-6}{2*-1}=\frac{-4}{-2}
=+2
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+6}{2*-1}=\frac{8}{-2}
=-4
$