Let's solve the quadratic equation:

$3x^2 - 4x = 4$

Step 1: Bring all terms to one side of the equation.

Subtract 4 from both sides:

$3x^2 - 4x - 4 = 0$

Step 2: Apply the quadratic formula.

The quadratic formula is:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

For the equation $3x^2 - 4x - 4 = 0$, we have:

• $a = 3$
• $b = -4$
• $c = -4$

Step 3: Plug in the values.

First, calculate the discriminant:

$\Delta = b^2 - 4ac = (-4)^2 - 4(3)(-4) = 16 + 48 = 64$

Now, solve for $x$:

$x = \frac{-(-4) \pm \sqrt{64}}{2(3)} = \frac{4 \pm 8}{6}$

Step 4: Find the two solutions.

• For $x_1$:

$x_1 = \frac{4 + 8}{6} = \frac{12}{6} = 2$

• For $x_2$:

$x_2 = \frac{4 - 8}{6} = \frac{-4}{6} = -\frac{2}{3}$

Final solutions:

The solutions are:

$x_1 = 2, \quad x_2 = -\frac{2}{3}$

Would you like further details or have any questions?

Here are five related questions:

1. How is the discriminant used to determine the nature of roots?
2. What are the steps to factor quadratic equations if possible?
3. Can quadratic equations always be solved using the quadratic formula?
4. How can we verify solutions by plugging them back into the original equation?
5. What happens if the discriminant is negative?

Tip: Always simplify the equation as much as possible before applying the quadratic formula.