Solve each of the following quadratic equations:
`2x^(2)-x+(1)/(8)=0`

Explanation

To solve the quadratic equation \(2x^2 - x + \frac{1}{8} = 0\), we will follow these steps:

### Step 1: Eliminate the Fraction
To eliminate the fraction, we can multiply the entire equation by 8 (the denominator).

\[ 8(2x^2) - 8(x) + 8\left(\frac{1}{8}\right) = 0 \]

This simplifies to:

\[ 16x^2 - 8x + 1 = 0 \]

### Step 2: Identify Coefficients
Now, we can identify the coefficients \(a\), \(b\), and \(c\) from the standard form of a quadratic equation \(ax^2 + bx + c = 0\).

Here, we have:
- \(a = 16\)
- \(b = -8\)
- \(c = 1\)

### Step 3: Use the Quadratic Formula
We can use the quadratic formula to find the roots of the equation, which is given by:

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

### Step 4: Calculate the Discriminant
First, we calculate the discriminant \(D = b^2 - 4ac\):

\[ D = (-8)^2 - 4(16)(1) = 64 - 64 = 0 \]

### Step 5: Calculate the Roots
Since the discriminant is 0, we have real and equal roots. Now, substituting the values into the quadratic formula:

\[ x = \frac{-(-8) \pm \sqrt{0}}{2(16)} = \frac{8 \pm 0}{32} = \frac{8}{32} = \frac{1}{4} \]

### Step 6: Conclusion
Thus, the roots of the equation are:

\[ x = \frac{1}{4} \quad \text{(with multiplicity 2)} \]