To find $(f = g)(x)$, we need to solve the equation where $f(x)$ equals $g(x)$. This means we will set the two functions equal to each other and solve for $x$.

Given: $f(x) = x^2 - x - 3$ $g(x) = 5x - 2$

Set $f(x)$ equal to $g(x)$: $x^2 - x - 3 = 5x - 2$

Now, let's bring all the terms to one side of the equation: $x^2 - x - 3 - 5x + 2 = 0$

Combine like terms: $x^2 - 6x - 1 = 0$

This is a quadratic equation. We can solve it using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 1$, $b = -6$, and $c = -1$.

Substituting the values: $x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(-1)}}{2(1)}$ $x = \frac{6 \pm \sqrt{36 + 4}}{2}$ $x = \frac{6 \pm \sqrt{40}}{2}$ $x = \frac{6 \pm 2\sqrt{10}}{2}$ $x = 3 \pm \sqrt{10}$

So, the solutions are: $x = 3 + \sqrt{10} \quad \text{or} \quad x = 3 - \sqrt{10}$

These are the points where $f(x) = g(x)$.

Would you like further details or have any questions?

Here are 5 related questions:

1. How do you determine the vertex of the parabola $f(x) = x^2 - x - 3$?
2. What is the significance of the discriminant in solving quadratic equations?
3. How do you find the points of intersection between two functions graphically?
4. What are the roots of the quadratic function $f(x) = x^2 - 6x - 1$?
5. Can the quadratic equation always be solved by factoring?

Tip: When solving quadratic equations, always check if it can be factored before using the quadratic formula, as factoring can be quicker and simpler.