How do you factor the trinomial $x^2-5x-45=0$ ?

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Answer 1 · 2016-04-12T12:26:58

not factorable.

$y = x^2 - 5x - 45$
$D = b^2 - 4ac = 25 + 180 = 205.$
Since D is not a perfect square, then this trinomial can't be factored

How do you factor the trinomial x^2-5x-45=0x^2-5x-45=0?