Answer
x = -1 or x = 5
Work Step by Step
$x^{2}-4x-5=0$
Part A: By Factoring
1) Factor the equation:
$(x-5)(x+1) = 0$
2) Using the Zero-Product Property we get:
$x-5 = 0$ or $x+1 = 0$
3) Solve the two simpler equations:
$x = 5$ or $x = -1$
Part B: By Completing the Squares
1) To make it a perfect square we need to first move the (-5) to the other side by adding 5 to each side.
$x^{2}-4x = 5$
2) Then we make it square by adding 4 to each side.
$x^{2}-4x+4 = 9$
3) We can now factor that giving us:
$(x-2)^{2} = 9$
4) Get the square root of each side:
$x-2 = \frac{+}{-}\sqrt 9$
$x-2 = 3$ or $x-2 = -3$
5) Solve the simple equations:
$x = 5$ or $x = -1$
Part C: Using the Quadratic Formula
$x=\frac{-b\frac{+}{-}\sqrt (b^{2}-4ac)}{2a}$
1) Plug the equation into the formula:
$x=\frac{-(-4)\frac{+}{-}\sqrt ((-4)^{2}-4(1)(-5))}{2(1)}$
2) Solve for x:
$x=\frac{4\frac{+}{-}\sqrt (16+20)}{2}$
$x=\frac{4\frac{+}{-}\sqrt (36)}{2}$
$x=\frac{4\frac{+}{-}6}{2}$
$x=\frac{10}{2}$ or $x=\frac{-2}{2}$
$x=5$ or $x=-1$
