To find the values of $x$ where the curves $y = 4x^2 + 6x - 4$ and $y = 2x + 1$ intersect, we need to solve for $x$ when the two equations are equal.

1. Set the two equations equal to each other: $4x^2 + 6x - 4 = 2x + 1$

2. Rearrange the equation to move all terms to one side: $4x^2 + 6x - 4 - 2x - 1 = 0$ $4x^2 + 4x - 5 = 0$

3. Solve the quadratic equation: $4x^2 + 4x - 5 = 0$

We can apply the quadratic formula. For a quadratic equation $ax^2 + bx + c = 0$, the solutions are given by: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here, $a = 4$, $b = 4$, and $c = -5$.

Substitute the values into the quadratic formula: $x = \frac{-4 \pm \sqrt{4^2 - 4(4)(-5)}}{2(4)}$ $x = \frac{-4 \pm \sqrt{16 + 80}}{8}$ $x = \frac{-4 \pm \sqrt{96}}{8}$ $x = \frac{-4 \pm 4\sqrt{6}}{8}$ $x = \frac{-1 \pm \sqrt{6}}{2}$

So the two solutions for $x$ are: $x = \frac{-1 + \sqrt{6}}{2} \quad \text{or} \quad x = \frac{-1 - \sqrt{6}}{2}$

These are the values of $x$ where the curves intersect.

Would you like further details or explanations?

Here are five related questions:

1. How can we find the corresponding $y$-values for the points of intersection?
2. What is the geometric meaning of finding the intersection of two curves?
3. How does the discriminant help in determining the number of intersection points?
4. How would the solution change if we had different coefficients in the quadratic equation?
5. Can these curves have more than two intersection points?

Tip: Always simplify your equation before applying the quadratic formula for easier calculations.