The given equation is:

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

This is a quadratic equation in the form of $ax^2 + bx + c = 0$, where $a = 4$, $b = 4$, and $c = 1$.

To solve it, we can apply the quadratic formula:

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

Substituting the values $a = 4$, $b = 4$, and $c = 1$ into the formula:

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

$x = \frac{-4 \pm \sqrt{16 - 16}}{8}$

$x = \frac{-4 \pm \sqrt{0}}{8}$

$x = \frac{-4 \pm 0}{8}$

$x = \frac{-4}{8}$

$x = -\frac{1}{2}$

So, the solution is:

$x = -\frac{1}{2}$

This is a repeated real root (the equation has only one solution).

Would you like to explore more about the solution, or is there anything else you'd like to ask?

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Here are some related questions:

1. What is the discriminant of a quadratic equation, and how does it determine the nature of the roots?
2. Can you explain how to factor a quadratic equation like this?
3. How do you identify whether a quadratic equation has real, imaginary, or complex roots?
4. What are the possible outcomes when solving quadratic equations using the quadratic formula?
5. What happens when the discriminant of a quadratic equation equals zero?

Tip: When the discriminant ($b^2 - 4ac$) of a quadratic equation equals zero, the equation has exactly one real solution, which is also known as a double root.