To solve the quadratic equation \( 2x^2 + 3x + 1 = 0 \), we can use the quadratic formula:

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

Here, \( a = 2 \), \( b = 3 \), and \( c = 1 \). Plugging these values into the formula, we get:

\[
x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot 1}}{2 \cdot 2}
\]

First, calculate the discriminant:

\[
\Delta = b^2 - 4ac = 3^2 - 4 \cdot 2 \cdot 1 = 9 - 8 = 1
\]

Since the discriminant is positive, we have two real solutions:

\[
x = \frac{-3 \pm \sqrt{1}}{4}
\]

Now, compute the two solutions:

\[
x = \frac{-3 + 1}{4} = \frac{-2}{4} = -\frac{1}{2}
\]

\[
x = \frac{-3 - 1}{4} = \frac{-4}{4} = -1
\]

So, the solutions to the equation \( 2x^2 + 3x + 1 = 0 \) are \( x = -\frac{1}{2} \) and \( x = -1 \).

Would you like more details or have any questions?

Here are five more questions you might consider asking:
1. How do you solve the quadratic equation \( x^2 - 4x + 4 = 0 \)?
2. Can you explain how to complete the square for the equation \( x^2 + 6x + 5 = 0 \)?
3. What are the solutions to the equation \( 3x^2 - x - 2 = 0 \)?
4. How do you graph the quadratic function \( y = 2x^2 + 3x + 1 \)?
5. What is the vertex form of a quadratic equation?

**Tip:** Always check the discriminant (\( b^2 - 4ac \)) first when solving a quadratic equation to determine the nature of the roots (real and distinct, real and repeated, or complex).

Learn how to solve the quadratic equation 2x^2 + 3x + 1 = 0 with a detailed step-by-step solution. This problem covers key algebraic concepts, including the quadratic formula and discriminant. Suitable for high school students.

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