Use the substitution method to determine whether the linear-quadratic system y=x2−2x−6 , y=−5x−9 has 0, 1, or 2 solutions.(1 point)

To solve the system of equations using the substitution method, we will substitute the expression for \( y \) from the second equation into the first equation.

The given equations are:

1. \( y = x^2 - 2x - 6 \)
2. \( y = -5x - 9 \)

We can set the two equations for \( y \) equal to each other:

\[
x^2 - 2x - 6 = -5x - 9
\]

Next, we rearrange this equation to bring all terms to one side:

\[
x^2 - 2x + 5x - 6 + 9 = 0
\]

Combining like terms gives:

\[
x^2 + 3x + 3 = 0
\]

Now, we want to determine the number of solutions to this quadratic equation. We can do this by calculating the discriminant \( D \):

\[
D = b^2 - 4ac
\]

Here, \( a = 1 \), \( b = 3 \), and \( c = 3 \):

\[
D = 3^2 - 4(1)(3) = 9 - 12 = -3
\]

Since the discriminant \( D \) is less than 0, this quadratic equation has no real solutions. Hence, the system of equations has no solutions.

The final answer is:

This system has **0 solution(s)**.