Is (–8, 0) a solution to the equation y = x2 − 5?

To determine if the point \((-8, 0)\) is a solution to the equation \(y = x^2 - 5\), we can substitute \(x = -8\) into the equation and see if we get \(y = 0\).

Starting with the equation:

\[
y = x^2 - 5
\]

Substituting \(x = -8\):

\[
y = (-8)^2 - 5
\]

Calculating \((-8)^2\):

\[
(-8)^2 = 64
\]

Now substituting back into the equation:

\[
y = 64 - 5 = 59
\]

Now we have \(y = 59\) when \(x = -8\). The point \((-8, 0)\) gives \(y = 0\), but we've determined that \(y = 59\).

Since \(0 \neq 59\), the point \((-8, 0)\) is **not** a solution to the equation \(y = x^2 - 5\).