To find the points of intersection of the two given equations:

1. $y = 2x^2 + x - 1$
2. $2y - 3x = 58$

we solve them simultaneously.

Step 1: Rearrange the second equation

Rearranging $2y - 3x = 58$ for $y$: $2y = 3x + 58$ $y = \frac{3x}{2} + 29$

Step 2: Substitute $y$ into the first equation

Substitute $y = \frac{3x}{2} + 29$ into $y = 2x^2 + x - 1$: $\frac{3x}{2} + 29 = 2x^2 + x - 1$

Multiply through by 2 to eliminate the fraction: $3x + 58 = 4x^2 + 2x - 2$

Simplify: $4x^2 - x - 60 = 0$

Step 3: Solve the quadratic equation

The quadratic equation is $4x^2 - x - 60 = 0$. Using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 4$, $b = -1$, $c = -60$.

$x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(4)(-60)}}{2(4)}$ $x = \frac{1 \pm \sqrt{1 + 960}}{8}$ $x = \frac{1 \pm \sqrt{961}}{8}$ $x = \frac{1 \pm 31}{8}$

So: $x = \frac{32}{8} = 4 \quad \text{or} \quad x = \frac{-30}{8} = -\frac{15}{4}.$

Step 4: Find corresponding $y$-values

For $x = 4$, substitute into $y = \frac{3x}{2} + 29$: $y = \frac{3(4)}{2} + 29 = 6 + 29 = 35.$

For $x = -\frac{15}{4}$, substitute into $y = \frac{3x}{2} + 29$: $y = \frac{3(-\frac{15}{4})}{2} + 29 = -\frac{45}{8} + 29 = \frac{-45 + 232}{8} = \frac{187}{8}.$

Final Answer:

The points of intersection are: $(4, 35) \quad \text{and} \quad \left(-\frac{15}{4}, \frac{187}{8}\right).$

Would you like detailed steps for the quadratic formula or simplification?

Here are some related questions to explore:

1. How do we confirm these points are correct graphically?
2. What methods other than substitution can solve simultaneous equations?
3. How does the shape of the parabola affect the number of intersection points?
4. Can this problem be solved numerically or by approximation?
5. What role does the quadratic discriminant play in determining the solutions?

Tip: Always check your solutions by substituting them back into both original equations.