How do you solve #x² + x = 6#?

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How do you solve #x² + x = 6#?

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Answer 1 · 2016-03-24T14:30:10

#x=-3, 2#

Explanation:

#x^2+x=6# => Subtract 6 from both sides:
#x^2+x-6=0# => #x=3x-2x#:
#x^2+3x-2x-6=0# => factor by grouping:
#x(x+3)-2(x+3)=0#
#(x+3)(x-2)=0#
The Zero Product Property: if ab=0, then either a=0 or b=0, or both.
#x+3=0# => #x=-3#
#x-2=0# => #x=2#

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How do you solve #x² + x = 6#?

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