The problem asks to find the intersection points of the curve $y = 2x^2 + 3x - 2$ and the line $y = 2x - 1$.

1. Problem Description

The problem asks to find the intersection points of the curve $y = 2x^2 + 3x - 2$ and the line $y = 2x - 1$.

2. Solution Steps

To find the intersection points, we need to solve the system of equations:

$y = 2x^2 + 3x - 2$

$y = 2x - 1$

We can substitute the second equation into the first equation:

$2x - 1 = 2x^2 + 3x - 2$

Now we rearrange the equation to get a quadratic equation in standard form:

$0 = 2x^2 + 3x - 2 - 2x + 1$

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

We can solve this quadratic equation by factoring:

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

$2x(x + 1) - 1(x + 1) = 0$

$(2x - 1)(x + 1) = 0$

This gives us two possible values for $x$:

$2x - 1 = 0 \Rightarrow 2x = 1 \Rightarrow x = \frac{1}{2}$

$x + 1 = 0 \Rightarrow x = -1$

Now we find the corresponding $y$ values for each $x$ value by substituting them into the equation $y = 2x - 1$:

For $x = \frac{1}{2}$:

$y = 2(\frac{1}{2}) - 1 = 1 - 1 = 0$

So, one intersection point is $(\frac{1}{2}, 0)$.

For $x = -1$:

$y = 2(-1) - 1 = -2 - 1 = -3$

So, the other intersection point is $(-1, -3)$.

3. Final Answer

The intersection points are $(\frac{1}{2}, 0)$ and $(-1, -3)$.