How do you solve $(x + 2) (x - 4) = 12$ $(x+2)(x-4)=12$ ?

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How do you solve $(x + 2) (x - 4) = 12$ $(x+2)(x-4)=12$ ?

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Answer 1 · 2017-10-11T06:27:21

See the answer below...

Explanation:

$(x + 2) (x - 4) = 12$ $(x+2)(x-4)=12$
$\Rightarrow x^{2} - 2 x - 8 = 12$ $=>x^2-2x-8=12$
$\Rightarrow x^{2} - 2 x - 20 = 0$ $=>x^2-2x-20=0$

Now apply Pythagoras rule,
The answt is following....
![my http://pleasure....](https://useruploads.socratic.org/oJTkMDHTJm6qKznj6xY4_Screenshot_2017-10-11-11-54-45.png)

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How do you solve $(x + 2) (x - 4) = 12$ $(x+2)(x-4)=12$ ?

1 Answer

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How do you solve (x+2)(x−4)=12(x+2)(x-4)=12?

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Explanation:

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How do you solve $(x + 2) (x - 4) = 12$ $(x+2)(x-4)=12$ ?