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

Question

1281 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-20T07:16:22

$z = - 7 and \frac{3}{2}$ $z = -7 and 3/2$

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

$4 z^{2} + 22 z - 42 = 0$ $4z^2 + 22z - 42 = 0$

$2 (2 z^{2} + 11 z - 21) = 0$ $2(2z^2 + 11z - 21) = 0$

$2 z^{2} + 11 z - 21 = 0$ $2z^2 + 11z - 21 = 0$

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

$= 2 z^{2} - 3 z + 14 z - 21 = 0$ $= 2z^2 - 3z + 14z - 21 = 0$

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

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

$\Rightarrow z + 7 = 0 and 2 z - 3 = 0$ $=> z + 7 = 0 and 2z-3 = 0$

$\Rightarrow z = - 7 and \frac{3}{2}$ $=> z = -7 and 3/2$

Or you could use quadratic formula!

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