Displacement of spring of a block going down an incline

[joedango]

Homework Statement

An inclined plane of angle θ = 20.0° has a spring of force constant k = 455 N/m fastened securely at the bottom so that the spring is parallel to the surface as shown in the figure below. A block of mass m = 2.71 kg is placed on the plane at a distance d = 0.330 m from the spring. From this position, the block is projected downward toward the spring with speed v = 0.750 m/s. By what distance is the spring compressed when the block momentarily comes to rest?

Homework Equations

dE=dk+du=0

Kf-Ki+Dfsp-Disp+Dfg-Dig=0
Ki=1/2mv^2
Dfsp=1/2kx^2
Dig= mg[h + x(sin20.0)]

The Attempt at a Solution

Kf,Disp, and Dfg all Equal zero
so by substitution:

-1/2mv^2+1/2kx^2-mg[h + x(sin20.0)]=0
-(2.71)(.75^2)/2+455/2x^2-(2.71*9.8)[0.33+xsin(20)]=0

which simplifies to
227.5x^2-9.084x-9.52=0

My answer is .226, which is the incorrect answer. What am I doing wrong?

Thanks!

 

[joedango]

I forgot to post the picture. Here it is: