The equation given is:

$6x^2 - 4x - 1 = 0$

We are tasked with finding the complex solutions of this quadratic equation. To solve it, we will use the quadratic formula:

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

Here, the coefficients from the equation $6x^2 - 4x - 1 = 0$ are:

• $a = 6$
• $b = -4$
• $c = -1$

Step 1: Calculate the discriminant

The discriminant $\Delta$ is given by:

$\Delta = b^2 - 4ac$

Substituting the values:

$\Delta = (-4)^2 - 4(6)(-1) = 16 + 24 = 40$

Step 2: Apply the quadratic formula

Now substitute the values of $a$, $b$, and $\Delta$ into the quadratic formula:

$x = \frac{-(-4) \pm \sqrt{40}}{2(6)} = \frac{4 \pm \sqrt{40}}{12}$

Step 3: Simplify the square root

The square root of 40 can be simplified as:

$\sqrt{40} = \sqrt{4 \times 10} = 2\sqrt{10}$

Thus, the expression for $x$ becomes:

$x = \frac{4 \pm 2\sqrt{10}}{12}$

Step 4: Simplify the fractions

Now simplify the fraction:

$x = \frac{4}{12} \pm \frac{2\sqrt{10}}{12} = \frac{1}{3} \pm \frac{\sqrt{10}}{6}$

Final Solutions

So, the two complex solutions for $x$ are:

$x_1 = \frac{1}{3} + \frac{\sqrt{10}}{6}, \quad x_2 = \frac{1}{3} - \frac{\sqrt{10}}{6}$

These are the solutions in terms of real numbers, but since the discriminant is positive, they are both real, not complex.

Would you like further clarification or details?

Here are five related questions:

1. How does the discriminant affect the nature of solutions (real vs complex)?
2. Can you always use the quadratic formula to solve any quadratic equation?
3. What happens if the discriminant is negative?
4. How can you solve quadratics using factoring instead of the quadratic formula?
5. How does completing the square compare to using the quadratic formula?

Tip: Always simplify the square root and fractions when solving quadratic equations to get the cleanest form of the solution.