Trouble understanding the influence of a real spring in a system

[hcpyz]

The hypothesis is that the force of a real spring can be described as $$F = -kx + \alpha x^2$$ with x being the spring deformation and k its constant. The \alpha x^2 would be the force lost by the spring as x becomes too big. To test that, a system was build with block of mass m suspended by a vertical spring. Different blocks with different masses were used and, in the equilibrium state, the value y of its height in relation to the ground was registered. Worth mentioning that for the problem $$x = y_0 - y$$ with y_0 being the height of the lower end of the spring in relation to the ground when there is no block.

**The problem is**: Find a function y = y(m) in terms of k, alpha and y_0 .

**What I've done so far**

I thought that since we are dealing with a mass spring system we would find an equation for the system without any mass being weighted. So:
$$
-k(y_0) + \alpha(y_0)^2 = 0
$$
I don't feel very confident with that statement but moving on...
The I found an equation for the system when there's a mass being weighted, which would be found by doing the force diagram. This is being really confusing to me, so...

$$
P = mg = -kx + \alpha x^2
$$
Here I'm really confused with the minus sing. But well, moving on again...
$$
mg = -k(y_0 - y) + \alpha (y_0 - y)^2
$$
And then I thought that it would be a good idea to solve this system.

$$
\begin{cases}
-k(y_0) + \alpha(y_0)^2 = 0 \\
mg = -k(y_0 - y) + \alpha (y_0 - y)^2
\end{cases}
$$

I've tried doing it by hand and also with wolfram mathematica but it seems unsolvable thus I think I'm missing the conceptual part of it in the force diagram I built with the equation, am I getting the minus sing wrong?

 

[Lnewqban]

Welcome!
Do you know the type of deformation the spring is suffering: compression, torsion or traction?
That equation is difficult for me to translate to a real situation.

Please, see the second diagram in this link:
https://courses.lumenlearning.com/boundless-physics/chapter/hookes-law/

 

[kuruman]

If $~mg = -k(y_0 - y) + \alpha (y_0 - y)^2~,$ what prevents you from solving the quadratic for $(y_0-y)$ and then moving $y_0$ to the other side of the equation? Wouldn't that give you $y(m)$? Be sure to choose the solution that is physically meaningful.

Addendum on edit:
The "no mass" equation you have is incorrect. The variable $y$ is the height of the mass above ground. We are also given that the height above ground when no mass is hanging is $y_0$. So if you substitute $m=0$ and $y=y_0$ in the force equation, you get $$0*g=-k(y_0-y_0)+\alpha(y_0-y_0)^2\implies0=0.$$Nothing usable here.

 

[hcpyz]

If $~mg = -k(y_0 - y) + \alpha (y_0 - y)^2~,$ what prevents you from solving the quadratic for $(y_0-y)$ and then moving $y_0$ to the other side of the equation? Wouldn't that give you $y(m)$? Be sure to choose the solution that is physically meaningful.

Addendum on edit:
The "no mass" equation you have is incorrect. The variable $y$ is the height of the mass above ground. We are also given that the height above ground when no mass is hanging is $y_0$. So if you substitute $m=0$ and $y=y_0$ in the force equation, you get $$0*g=-k(y_0-y_0)+\alpha(y_0-y_0)^2\implies0=0.$$Nothing usable here.

oh wow, that was way simpler than I was thinking. Thank you!