Let f be the piecewise function such that

f(x) = { x2-5x-64 if x<=0,

{ x2-6x+10 if x>0

At how many points x does f(x)=2?

I got 2 before, but it says that I'm wrong.

x^{2}-5x-64=2 ⇒ x^{2}-5x-66=0 ⇒ (x-11)(x+6)=0. x = -6 is the only negative solution.

x^{2}-6x+10=2 ⇒ x^{2}-6x+8=0 ⇒ (x-4)(x-2)=0. x = 4 and x = 2 are both positive solutions.

Hence three solutions in total.