function or relation solve f(x)=2x^2-3x+1
show your solutions

To solve \( f(x) = 2x^2 - 3x + 1 \), we interpret this as a quadratic equation and find the values of \( x \) that satisfy \( f(x) = 0 \).

### Step 1: Set the function equal to zero

We set the equation \( f(x) = 2x^2 - 3x + 1 \) to zero:

\[
2x^2 - 3x + 1 = 0
\]

### Step 2: Use the quadratic formula

For any quadratic equation of the form \( ax^2 + bx + c = 0 \), the solutions for \( x \) are given by the quadratic formula:

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

In this case, the coefficients are:
- \( a = 2 \)
- \( b = -3 \)
- \( c = 1 \)

Now, substitute the values of \( a \), \( b \), and \( c \) into the quadratic formula:

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

### Step 3: Simplify the expression

First, calculate the discriminant \( \Delta \):

\[
\Delta = (-3)^2 - 4(2)(1) = 9 - 8 = 1
\]

Now, use the discriminant to find the solutions:

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

### Step 4: Find the two possible solutions

1. \( x_1 = \frac{3 + 1}{4} = \frac{4}{4} = 1 \)
2. \( x_2 = \frac{3 - 1}{4} = \frac{2}{4} = \frac{1}{2} \)

### Final Answer:

The two solutions are:
\[
x = 1 \quad \text{and} \quad x = \frac{1}{2}
\]

Would you like more details, or do you have any questions?

Here are some additional questions you might want to explore:

1. How do we determine the discriminant's significance in quadratic equations?
2. What is the role of the quadratic formula in solving equations?
3. Can a quadratic equation have no real solutions? If so, when?
4. How can we graph the quadratic function \( f(x) = 2x^2 - 3x + 1 \)?
5. What are some applications of quadratic equations in real-world problems?

**Tip:** Always check the discriminant value \( b^2 - 4ac \) before solving to know the nature of the roots.

function or relation solve f(x)=2x^2-3x+1 show your solutions

Learn how to solve the quadratic equation f(x) = 2x^2 - 3x + 1 with a step-by-step solution. This detailed guide explains the use of the quadratic formula, focusing on key algebraic concepts for students in Grades 8-10.

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