How do you solve #u^ { 2} - 6u - 27#?

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Answer 1 · 2017-02-25T02:10:24

#u=9# and #u=-3#

#u^2-6u-27# can be written as:

#(u-9)(u+3)=0#

Solve for #u#. If one of the factors gives a value of #0#, then #u# is #0# at that spot.

#u-9=0#

#u=9#

#u+3=0#

#u=-3#

You can check this graphically:

graph{x^2-6x-27 [-10.18, 15.13, -5.87, 6.79]}