Solve x for a quadratic equation (not finding zeroes) With a linear function $f(x)=5x+2=q$ 
can be solved for $x$ by rewriting it as
$x=(q-2)/5$ 
While with a quadratic function 
$f(x)=5x^2+2x+2=q$ how would you solve for both x's on one side?
So you clearly can not add the solutions for $5x^2=q$ and $2x+2=q$.
Is there an equation to relate y values of $5x^2=$(some relation of q) and $2x+2=$(some relation of q) to q? 
 A: Your idea is quite exceptional. It is not generally adopted because the values of x in a quadratic equation can readily be found by methods like factorization, completing square and as well as quadratic formula.
Your suggestion is to break the original into P: 2x + 2 = k and Q : 5x^2 = q – k and hope that the k (if can easily be found) can help in solving the equation with lesser effort. Let see if that is possible or not.
By eliminating x from P and Q, we end up with $5k^2 – 16k + (20 – 4q) = 0$.
Of course, the value(s) of k can be found but we need to solve another quadratic! The worst thing has yet to be come – extraneous value of k must be distinguished and be discarded.
The following uses q = 5 as an example:-
By ‘normal’ methods, we found x = –1 or 3/5.
Using your approach, we have … k = 0 or 16/5.
When k = 0, from P, we get x = –1, but from Q, we get x =–1 or 1(has to be rejected)
When k = 16/5, from P, x = 3/5, but from Q, we get x = 3/5 or –3/5 (has to be rejected)
Note also that the above is just an exceptionally simple example. That means not every quadratic in k has integral/fractional/real roots.
Summing up:- Your approach is a possible way to solve a quadratic but time is wasted in doing many extra work.
A: So you are asking us to find two convenient functions $f,g$ such that$$5x^2=f(q),\\2x+2=g(q),$$ which are related by $$f(q)+g(q)=q.$$
Obviously, we can eliminate $g(q):=q-f(q)$, and
$$5x^2=f(q),\\2x+2=q-f(q).$$
This system is compatible iff
$$\frac{f(q)}5=\left(\frac{q-f(q)-2}2\right)^2.$$
The solution of this quadratic equation are
$$f(q)=q\pm2\sqrt{\frac q5-\frac9{25}}-\frac85,$$
implying
$$g(q)=\mp2\sqrt{\frac q5-\frac9{25}}+\frac85.$$
So yes, it is possible to find these functions, they are well defined, but they don't help making things simpler, on the opposite.

The way that works is by completing the square
$$5x^2+2x+2=5(x+\frac15)^2+2-5\left(\frac15\right)^2=q,$$
then
$$\left(x+\frac15\right)^2=\frac q5-\frac9{25}.$$
