Let's solve the system of equations using substitution. The system is:

y = -2x - 5 \tag{1} -3x^2 + 4x - y = 8 \tag{2}

Step 1: Substitute $y$ from equation (1) into equation (2).

Substitute $y = -2x - 5$ into equation (2):

$-3x^2 + 4x - (-2x - 5) = 8$

Simplify:

$-3x^2 + 4x + 2x + 5 = 8$ $-3x^2 + 6x + 5 = 8$

Step 2: Solve for $x$.

Subtract 8 from both sides:

$-3x^2 + 6x + 5 - 8 = 0$ $-3x^2 + 6x - 3 = 0$

Divide the entire equation by -3 to simplify:

$x^2 - 2x + 1 = 0$

This simplifies to:

$(x - 1)^2 = 0$

So, $x = 1$.

Step 3: Substitute $x = 1$ into equation (1) to find $y$.

Using $y = -2x - 5$:

$y = -2(1) - 5 = -2 - 5 = -7$

Thus, the solution is $(x, y) = (1, -7)$.

Final Answer:

The solution is $\boxed{(1, -7)}$.

Would you like further details on any part of the solution?

Here are 5 related questions you might find helpful:

1. How do you solve quadratic equations using factoring?
2. What are other methods to solve a system of equations, like elimination?
3. How do you check if a solution is correct for a system of equations?
4. What are some real-world applications of solving systems of equations?
5. How do you solve a quadratic equation when it doesn't factor easily?

Tip: Always double-check your solution by plugging the values back into the original equations!