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!

How do you solve 2z ( 2z + 11) = 422z ( 2z + 11) = 42?