To solve the quadratic equation:

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

We will use the quadratic formula:

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

For the given equation, the coefficients are:

• $a = -4$
• $b = 6$
• $c = 1$

Now, substituting these values into the quadratic formula:

$x = \frac{-6 \pm \sqrt{6^2 - 4(-4)(1)}}{2(-4)}$

First, calculate the discriminant:

$b^2 - 4ac = 6^2 - 4(-4)(1) = 36 + 16 = 52$

Now substitute this back into the formula:

$x = \frac{-6 \pm \sqrt{52}}{-8}$

Simplify $\sqrt{52}$:

$\sqrt{52} = \sqrt{4 \times 13} = 2\sqrt{13}$

Now substitute this into the formula:

$x = \frac{-6 \pm 2\sqrt{13}}{-8}$

We can simplify the expression by dividing each term by $-2$:

$x = \frac{3 \mp \sqrt{13}}{4}$

So the two possible solutions are:

$x_1 = \frac{3 + \sqrt{13}}{4} \quad \text{and} \quad x_2 = \frac{3 - \sqrt{13}}{4}$

These are the two solutions to the quadratic equation.

---

Would you like any further details or clarifications on this solution?

Here are some related questions you might find interesting:

1. How do you factor a quadratic equation like $-4x^2 + 6x + 1 = 0$?
2. What is the significance of the discriminant in a quadratic equation?
3. How can we solve quadratic equations using completing the square?
4. What happens if the discriminant is negative in a quadratic equation?
5. Can you solve cubic equations using similar methods?

Tip: When using the quadratic formula, always first check if the discriminant (the value under the square root) is positive, zero, or negative, as this determines whether the solutions are real or complex.