A 0.321-kg mass is attatched to a spring with force constant of 13.3 N/m. If the mass is displaced 0.256 m from the Equalibrium and released, what is its speed when it is 0.128 m from equalibrium?

I used the formula (1/2) mv^2 = (1/2) kd^2. I solved for k and got 6.44 rad/s and then i solved the equation for v. I then used. 128 for my d. I plugged in my answer and got 8.24. I also used d as. 256 and got 1.64