To solve the system of equations given by:
- \( y = -12x - 43 \)
- \( y = x^2 - 30x + 38 \)
we can set these two equations equal to each other since both are equal to \( y \):
\[ -12x - 43 = x^2 - 30x + 38 \]
Rearranging this equation gives us:
\[ 0 = x^2 - 30x + 38 + 12x + 43 \]
Combine the like terms:
\[ 0 = x^2 - 18x + 81 \]
Now, we can rearrange this into standard quadratic form:
\[ x^2 - 18x + 81 = 0 \]
Next, we can factor this quadratic:
\[ (x - 9)(x - 9) = (x - 9)^2 = 0 \]
Setting the factor equal to zero gives us:
\[ x - 9 = 0 \implies x = 9 \]
Now, we substitute \( x = 9 \) back into the first equation to find \( y \):
\[ y = -12(9) - 43 \]
Calculating the right-hand side:
\[ y = -108 - 43 = -151 \]
Thus, we have found one solution to the system of equations:
\[ (x, y) = (9, -151) \]
So the coordinates in exact form are:
\[ \boxed{(9, -151)} \]