Evaluate $\log_{5} x^2(y-3)$ Given that $\log _{5} x=-2$ and $\log_{2} y=3$

Evaluate $\log_{5} x^2(y-3)$

My attempt, 

$=\log{5} x^2+ \log_{5} (y-3)$

I'm stuck here. Can anyone explain how to proceed? Thanks in advance.
Going from the line $A =\log_{5}x^{2} + \log_{5}(y-3)$:

$= 2(-2) +\log_{5}(y-3)$

$\log_{2}y = 3 \Rightarrow y = 8$

$A = -4 +\log_{5}(8-3) = -4 + \log_{5}5 = -4 + 1 = -3$

Edit: note that $\log_{5}(y-3) \neq \log_{5}y -\log_{5}3$
HINT:

$\log _{5} x=-2 \implies x=5^{-2}$ and $\log_{2} y=3 \implies y=8$

Can you evaluate it now?