To solve the quadratic equation $x^2 + 4x - 5 = 0$, we can use the quadratic formula:

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

For the given equation $x^2 + 4x - 5 = 0$, the coefficients are:

• $a = 1$
• $b = 4$
• $c = -5$

Step 1: Plug the values into the quadratic formula:

$x = \frac{-4 \pm \sqrt{4^2 - 4(1)(-5)}}{2(1)}$

Step 2: Simplify inside the square root:

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

Step 3: Solve the square root:

$x = \frac{-4 \pm 6}{2}$

Step 4: Find the two possible values for $x$:

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

Thus, the solutions are: $x = 1 \quad \text{or} \quad x = -5$

Would you like more details on any step or have any questions?

Here are 5 related questions you might find helpful:

1. How do you derive the quadratic formula?
2. What are the conditions for a quadratic equation to have real solutions?
3. How do you complete the square for a quadratic equation?
4. Can all quadratic equations be factored, and how do you factor them?
5. How do you graph a quadratic equation?

Tip: Always check the discriminant $b^2 - 4ac$ to determine if the solutions are real, complex, or repeated!