Find the real​ solutions, if​ any, of the following equation. Use the quadratic formula.
49t^2-14t+1=0

To solve the quadratic equation \(49t^2 - 14t + 1 = 0\) using the quadratic formula, we follow these steps:

The quadratic formula is given by:

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

Here, the coefficients are:
- \( a = 49 \)
- \( b = -14 \)
- \( c = 1 \)

### Step 1: Calculate the Discriminant
The discriminant \(\Delta\) is calculated as follows:

\[
\Delta = b^2 - 4ac
\]

Substituting the values:

\[
\Delta = (-14)^2 - 4(49)(1)
\]

\[
\Delta = 196 - 196 = 0
\]

### Step 2: Calculate the Roots
Since the discriminant is 0, there is exactly one real solution. We now use the quadratic formula:

\[
t = \frac{-b \pm \sqrt{\Delta}}{2a}
\]

Substituting the values:

\[
t = \frac{-(-14) \pm \sqrt{0}}{2(49)}
\]

\[
t = \frac{14 \pm 0}{98} = \frac{14}{98} = \frac{1}{7}
\]

### Final Answer
The real solution is:

\[
t = \frac{1}{7}
\]

Would you like further details or have any questions about this solution? 

Here are 5 related questions:
1. How does the discriminant determine the nature of the roots?
2. What if the discriminant was negative, how would the solution change?
3. Can you solve the quadratic equation by completing the square instead?
4. How do you derive the quadratic formula?
5. How can quadratic equations be applied in real-world scenarios?

**Tip:** Always check the discriminant first to understand how many and what type of solutions to expect from a quadratic equation.

Learn how to solve the quadratic equation 49t^2 - 14t + 1 = 0 using the quadratic formula with a step-by-step solution. Understand the role of discriminant in determining the nature of roots. Suitable for high school students.

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