How do you solve $x^2-4x-5=0$ ?

Question

1167 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-04T14:39:56

See a solution process below:

First, factor the quadratic as:

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

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

Now, equate each term on the left side of the equation to $0$ and solve for $x$ :

Solution 1:

$x + 1 = 0$

$x + 1 - color(red)(1) = 0 - color(red)(1)$

$x + 0 = -1$

$x = -1$

Solution 2:

$x - 5 = 0$

$x - 5 + color(red)(5) = 0 + color(red)(5)$

$x - 0 = 5$

$x = 5$

The Solution Set Is:

$x = {-1, 5}$

How do you solve x^2-4x-5=0x^2-4x-5=0?