How do you solve #2z ( 2z + 11) = 42#?

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Answer 1 · 2017-11-20T07:16:22

# z = -7 and 3/2 #

#2z (2z+11) = 42#

#4z^2 + 22z - 42 = 0#

#2(2z^2 + 11z - 21) = 0#

# 2z^2 + 11z - 21 = 0 #

From here I'm going to split the middle term:

#= 2z^2 - 3z + 14z - 21 = 0 #

#= z (2z - 3) + 7 (2z - 3) = 0#

#= (z + 7) (2z - 3) = 0 #

#=> z + 7 = 0 and 2z-3 = 0 #

#=> z = -7 and 3/2 #

Or you could use quadratic formula!