$$ 1 \quad f ( x ) = x ^ { 2 } + 4 x - 5 $$

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We start with the function

$$
f(x) = x^2 + 4x - 5
$$

and set it equal to zero to find the roots:

$$
x^2 + 4x - 5 = 0
$$

To factor the quadratic, we look for two numbers that multiply to $$-5$$ and add up to $$4$$. The numbers $$-1$$ and $$5$$ satisfy these conditions because:

$$
-1 \times 5 = -5 \quad \text{and} \quad -1 + 5 = 4
$$

Thus, we can factor the quadratic as:

$$
(x - 1)(x + 5) = 0
$$

Now, set each factor equal to zero:

1. $$x - 1 = 0$$ gives
   $$
   x = 1
   $$

2. $$x + 5 = 0$$ gives
   $$
   x = -5
   $$

Therefore, the solutions are:

$$
x = 1 \quad \text{or} \quad x = -5
$$
The solutions are $$ x = 1 $$ or $$ x = -5 $$.

\( 1 \quad f ( x ) = x ^ { 2 } + 4 x - 5 \)

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