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Let's solve both equations using the quadratic formula.

### a) \((x - 2)(x + 1) = 3x - 4\)

First, expand the left-hand side:

\[
(x - 2)(x + 1) = x^2 + x - 2x - 2 = x^2 - x - 2
\]

Now, the equation becomes:

\[
x^2 - x - 2 = 3x - 4
\]

Move all terms to one side of the equation:

\[
x^2 - x - 2 - 3x + 4 = 0
\]

Simplify:

\[
x^2 - 4x + 2 = 0
\]

Now, apply the quadratic formula:

\[
x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(2)}}{2(1)}
\]

Simplify the discriminant:

\[
x = \frac{4 \pm \sqrt{16 - 8}}{2} = \frac{4 \pm \sqrt{8}}{2}
\]

Simplify further:

\[
x = \frac{4 \pm 2\sqrt{2}}{2}
\]

Divide by 2:

\[
x = 2 \pm \sqrt{2}
\]

Thus, the solutions are:

\[
x = 2 + \sqrt{2}, \quad x = 2 - \sqrt{2}
\]

### b) \(2x^2 - 5x = -4\)

Move all terms to one side:

\[
2x^2 - 5x + 4 = 0
\]

Apply the quadratic formula:

\[
x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(4)}}{2(2)}
\]

Simplify the discriminant:

\[
x = \frac{5 \pm \sqrt{25 - 32}}{4} = \frac{5 \pm \sqrt{-7}}{4}
\]

Since the discriminant is negative, the solutions are complex. Therefore, the solutions are:

\[
x = \frac{5 \pm \sqrt{7}i}{4}
\]

These are complex numbers.

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Would you like more details or have any questions about these solutions?

Here are some related questions for practice:
1. What is the importance of the discriminant in the quadratic formula?
2. How do you simplify square roots in the quadratic formula?
3. Can you apply the quadratic formula to solve \(x^2 - 6x + 9 = 0\)?
4. What are the solutions to the equation \(3x^2 + 6x - 9 = 0\)?
5. How would you solve a quadratic equation if the discriminant is zero?

**Tip**: Always check the discriminant before solving, as it determines whether the solutions are real or complex.

Solve the following equations with exact answers using the quadratic formula:
a) (x - 2)(x + 1) = 3x - 4
b) 2x^2 - 5x = -4

Learn how to solve quadratic equations using the quadratic formula with exact answers. This problem involves expanding expressions, simplifying quadratic equations, and finding solutions for both real and complex numbers. Suitable for students in Grades 9-10.

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