Find the real solutions to the equations. Part: $$ 0 / 2 $$ Part 1 of 2 (a) $$ \frac{4 x^{2}-36}{x^{2}-x+4}=0 $$ The solution set is

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The solution process is as follows:

1. Solve the numerator:
   $$
   4x^2 - 36 = 0 \quad \Longrightarrow \quad 4(x^2 - 9)= 0
   $$
   This implies:
   $$
   x^2 - 9 = 0 \quad \Longrightarrow \quad (x-3)(x+3)=0
   $$
   Therefore, the candidate solutions are:
   $$
   x=3 \quad \text{or} \quad x=-3
   $$

2. Verify that these solutions do not make the denominator zero:
   The denominator is:
   $$
   x^2 - x + 4
   $$
   For $$ x=3 $$:
   $$
   3^2 - 3 + 4 = 9 - 3 + 4 = 10 \neq 0
   $$
   For $$ x=-3 $$:
   $$
   (-3)^2 - (-3) + 4 = 9 + 3 + 4 = 16 \neq 0
   $$

Since neither $$ x=3 $$ nor $$ x=-3 $$ makes the denominator zero, both are valid.

The solution set is:
$$
\{-3,\,3\}
$$
The solutions are $$ x = -3 $$ and $$ x = 3 $$.

Find the real solutions to the equations. Part: \( 0 / 2 \) Part 1 of 2 (a) \( \frac{4 x^{2}-36}{x^{2}-x+4}=0 \) The solution set is

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