https://file.aiquickdraw.com/mathbot/file/1731749694697lnz7mj1j.jpg

From the image provided, the mathematical expression appears to be:

\[
\frac{5 + \sqrt{(-5)^2 - 4 \cdot 1 \cdot 4}}{2 \cdot 1}
\]

This is the quadratic formula used to find the roots of a quadratic equation \( ax^2 + bx + c = 0 \), with coefficients \( a = 1 \), \( b = -5 \), and \( c = 4 \). Let's solve it step by step:

1. **Quadratic Formula**:
   \[
   x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
   \]

2. **Substitute values of \( a = 1 \), \( b = -5 \), and \( c = 4 \):**
   \[
   x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(4)}}{2(1)}
   \]
   Simplify:
   \[
   x = \frac{5 \pm \sqrt{25 - 16}}{2}
   \]

3. **Simplify the discriminant**:
   \[
   \sqrt{25 - 16} = \sqrt{9} = 3
   \]

4. **Calculate the two possible roots**:
   \[
   x = \frac{5 + 3}{2} \quad \text{and} \quad x = \frac{5 - 3}{2}
   \]

   For the first root:
   \[
   x = \frac{8}{2} = 4
   \]

   For the second root:
   \[
   x = \frac{2}{2} = 1
   \]

**Final Answer**:
The roots of the quadratic equation are:
\[
x = 4 \quad \text{and} \quad x = 1
\]

---

Let me know if you'd like further clarification or a breakdown of any specific step! Here are some related questions for practice:

1. What happens if the discriminant \( b^2 - 4ac \) is negative?
2. Can you write the quadratic equation whose roots are \( x = 4 \) and \( x = 1 \)?
3. How does the quadratic formula generalize for complex roots?
4. How would you solve this quadratic equation graphically?
5. Can you derive the quadratic formula starting from the equation \( ax^2 + bx + c = 0 \)?

**Tip**: Always simplify the discriminant first to determine the nature of the roots (real and distinct, real and equal, or complex).

Solve the quadratic equation using the given expression: \[\frac{5 + \sqrt{(-5)^2 - 4 \cdot 1 \cdot 4}}{2 \cdot 1}\]

Step-by-step solution of a quadratic equation using the quadratic formula. The given equation has coefficients a = 1, b = -5, and c = 4. Learn how to find the roots by applying the quadratic formula and simplifying the discriminant.

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