How do you solve $4x=x^2-45$ ?

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Answer 1 · 2016-05-15T19:47:25

$(x - 9)(x + 4)$

It is a quadratic, so make it equal to 0.

$x^2 - 4x - 45 =0$

Find the factors of 45 which subtract to give 4.
The signs in the bracket are different - the greater factor will be negative.

$9 xx 5 = 45 and 9 -5 = 4$

$(x - 9)(x + 4)$

How do you solve 4x=x^2-454x=x^2-45?