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

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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}#