To solve the equation \(\frac{3}{8} + \frac{1}{2}t = \frac{2}{t}\), follow these steps:

### Step 1: Eliminate the fractions
To eliminate the fractions, multiply every term by the least common denominator (LCD). The LCD for the fractions is \(8t\) (since 8 is the least common multiple of 8 and 2, and \(t\) is the variable):

\[
8t \cdot \left(\frac{3}{8}\right) + 8t \cdot \left(\frac{1}{2}t\right) = 8t \cdot \left(\frac{2}{t}\right)
\]

This simplifies to:

\[
3t + 4t^2 = 16
\]

### Step 2: Rearrange the equation
Move all terms to one side of the equation to form a quadratic equation:

\[
4t^2 + 3t - 16 = 0
\]

### Step 3: Solve the quadratic equation
We can solve the quadratic equation \(4t^2 + 3t - 16 = 0\) using the quadratic formula:

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

Here, \(a = 4\), \(b = 3\), and \(c = -16\). Plugging in these values:

\[
t = \frac{-3 \pm \sqrt{3^2 - 4(4)(-16)}}{2(4)}
\]

\[
t = \frac{-3 \pm \sqrt{9 + 256}}{8}
\]

\[
t = \frac{-3 \pm \sqrt{265}}{8}
\]

### Step 4: Simplify the solutions
The solutions are:

\[
t = \frac{-3 + \sqrt{265}}{8} \quad \text{or} \quad t = \frac{-3 - \sqrt{265}}{8}
\]

These are the two possible values for \(t\).

Would you like more details or have any questions? Here are some related problems you might find useful:

1. Solve \(\frac{2}{5} + \frac{1}{3}t = \frac{3}{t}\).
2. Solve \(5t^2 + 6t - 7 = 0\) using the quadratic formula.
3. Factor the quadratic equation \(2t^2 - 4t - 6 = 0\).
4. Simplify \(\frac{3}{4} + \frac{2}{5t} = \frac{6}{t}\).
5. Solve the equation \(\frac{5}{t} + \frac{4}{t^2} = 3\).
6. Solve \(4t^2 + 7t + 3 = 0\) by completing the square.
7. Find the roots of \(3x^2 - 5x + 2 = 0\).
8. Solve \(\frac{1}{x} + \frac{2}{x + 1} = \frac{3}{x + 2}\).

**Tip:** Always double-check your solutions by substituting them back into the original equation to ensure they satisfy the equation.

Learn how to solve the equation 3/8 + 1/2t = 2/t with detailed steps and explanations. This problem involves eliminating fractions, rearranging equations, and solving a quadratic equation. Suitable for high school students studying algebra and quadratic equations.

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