To determine the value of \( c \) that makes the left-hand side of the equation \( x^2 - 6x + c = 15 \) a perfect square trinomial, we focus on the expression \( x^2 - 6x + c \).
A quadratic expression \( x^2 + bx + c \) can be factored into a perfect square trinomial if \( c \) is equal to \(\left(\frac{b}{2}\right)^2\).
In this case, \( b = -6 \). Therefore, we calculate:
\[ \frac{b}{2} = \frac{-6}{2} = -3 \]
Next, we find \(\left(-3\right)^2\):
\[ \left(-3\right)^2 = 9 \]
Thus, for \( x^2 - 6x + c \) to be a perfect square trinomial, we need:
\[ c = 9 \]
The modified equation becomes:
\[ x^2 - 6x + 9 = 15 \]
So, the value of \( c \) that makes the left side a perfect square trinomial is:
\[ \boxed{9} \]