Calculate how far the spring is stretched from its position?

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A 7.5kg block is suspended at rest from a spring (k=45N/m) along a incline as shown below (where theta= 17°)
 

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Well I am not sure what to do next
 

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Also I know that F=mg so I get 35N=-45N/m(x)
 
Start with a more careful free-body diagram for the block: what are the forces on it?
(hint: gravity, spring, and only one other...)

[edit: you appear to have Fk pointing directly upwards ... if this is the force from the spring you need to take another look at the diagram: which direction is the spring pulling on the block?
Don't know what Fp is supposed to be. Force of a "pull"? What is pulling on the black?]
 
Well there is force of gravity pushing down on the block and normal force pushing up than then the the force of the blocks mass is pulling the spring while the spring resists the pull...so that's what I have there in my free-body diagram
 
You are very confused about forces.
You listed four and there are only three - so you have one too many.

"pushing up" is too vague: the direction of the normal force is "upwards perpendicular to the ramp".

the spring force points "up along the ramp".

"gravity" is what we call the pull that something has due to it's mass.
you don't need a separate force for it. It points directly down.

redraw your free body diagram to show the correct directions and forces.
 
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So I changed my free-body diagram. .but I still can't get the answer .48m
 

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