To solve the quadratic equation \( k^2 - 3k - 4 = 0 \), we can use the quadratic formula:
\[ k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In this equation, \( a = 1 \), \( b = -3 \), and \( c = -4 \).
First, we need to calculate the discriminant \( b^2 - 4ac \):
\[ b^2 - 4ac = (-3)^2 - 4(1)(-4) = 9 + 16 = 25 \]
Now we substitute back into the quadratic formula:
\[ k = \frac{-(-3) \pm \sqrt{25}}{2(1)} = \frac{3 \pm 5}{2} \]
This gives us two solutions:
- When using the plus sign:
\[ k = \frac{3 + 5}{2} = \frac{8}{2} = 4 \]
- When using the minus sign:
\[ k = \frac{3 - 5}{2} = \frac{-2}{2} = -1 \]
Thus, the solutions are
\[ k = 4, -1 \]
The final answer is:
\[ k = 4, -1 \]