Let's solve the given equations step-by-step:

a) Solve $x^2 - 5 = 4x$

1. Rearrange the equation to standard quadratic form: $x^2 - 4x - 5 = 0$

2. Use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here:

   • $a = 1$,
   • $b = -4$,
   • $c = -5$.

3. Substitute into the formula: $x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-5)}}{2(1)}$

   Simplify step-by-step: $x = \frac{4 \pm \sqrt{16 + 20}}{2}$ $x = \frac{4 \pm \sqrt{36}}{2}$

4. Simplify further: $x = \frac{4 \pm 6}{2}$

   Solve for both cases:

   • $x = \frac{4 + 6}{2} = \frac{10}{2} = 5$
   • $x = \frac{4 - 6}{2} = \frac{-2}{2} = -1$

5. Solution: $x = 5 \quad \text{or} \quad x = -1$

b) Solve $x^2 + \frac{1}{5}x - \frac{4}{5} = 0$

1. Use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here:

   • $a = 1$,
   • $b = \frac{1}{5}$,
   • $c = -\frac{4}{5}$.

2. Substitute into the formula: $x = \frac{-\left(\frac{1}{5}\right) \pm \sqrt{\left(\frac{1}{5}\right)^2 - 4(1)\left(-\frac{4}{5}\right)}}{2(1)}$

3. Simplify step-by-step:

   • First calculate the discriminant ($b^2 - 4ac$): $\left(\frac{1}{5}\right)^2 - 4 \cdot 1 \cdot \left(-\frac{4}{5}\right)$ $= \frac{1}{25} + \frac{16}{5}$ Convert $\frac{16}{5}$ to have a denominator of 25: $\frac{16}{5} = \frac{80}{25}$ So: $\frac{1}{25} + \frac{80}{25} = \frac{81}{25}$

4. Update the quadratic formula: $x = \frac{-\frac{1}{5} \pm \sqrt{\frac{81}{25}}}{2}$

   Simplify the square root: $\sqrt{\frac{81}{25}} = \frac{9}{5}$

   So: $x = \frac{-\frac{1}{5} \pm \frac{9}{5}}{2}$

5. Solve for both cases:

   • Case 1: $x = \frac{-\frac{1}{5} + \frac{9}{5}}{2}$: $x = \frac{\frac{8}{5}}{2} = \frac{8}{10} = \frac{4}{5}$

   • Case 2: $x = \frac{-\frac{1}{5} - \frac{9}{5}}{2}$: $x = \frac{-\frac{10}{5}}{2} = \frac{-10}{10} = -1$

6. Solution: $x = \frac{4}{5} \quad \text{or} \quad x = -1$

Final Solutions:

a) $x = 5$ or $x = -1$
b) $x = \frac{4}{5}$ or $x = -1$

Would you like a more detailed explanation for any step? Let me know! Here are 5 related questions:

1. How is the quadratic formula derived?
2. What are the steps to factorize a quadratic equation instead of using the formula?
3. How do you determine the nature of roots using the discriminant?
4. Can you solve a quadratic equation graphically?
5. How can the quadratic formula be applied to real-world problems?

Tip: Always check solutions by substituting the values back into the original equation.