How do you solve #x^2 - 12 = 4x#?

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Answer 1 · 2018-06-01T14:50:04

See a solution process below:

First, put the equation in standard form:

#x^2 - 12 - color(red)(4x) = 4x - color(red)(4x)#

#x^2- 4x - 12 = 0#

Next, factor the left side of the equation as:

#(x - 6)(x + 2) = 0#

Now, solve each term on the left for #0# to solve the problem:

Solution 1:

#x - 6 = 0#

#x - 6 + color(red)(6) = 0 + color(red)(6)#

#x - 0 = 6#

#x = 6#

Solution 2:

#x + 2 = 0#

#x + 2 - color(red)(2) = 0 - color(red)(2)#

#x + 0 = -2#

#x = -2#

The Solution Set Is:

#x = {-2, 6}#