How do you solve #z^2-6z-16=0#?

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Answer 1 · 2017-09-11T02:21:14

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To factor determine what factors of #16# (#+-1#, #+-2#, #+-4#, #+-8#) add up to #-6#:

#(z - 8)(z + 2) = 0#

Now, solve each term for #0# to find the solutions for #x#:

Solution 1:

#z - 8 = 0#

#z - 8 + color(red)(8) = 0 + color(red)(8)#

#z - 0 = 8#

#z = 8#

Solution 2:

#z + 2 = 0#

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

#z + 0 = -2#

#z = -2#