How do I solve n^2-5n-14=0 by factoring

asked Jun 10, 2015 7.4k views

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$n^2-5n-14=0 \\ n^2-7n+2n-14=0\\ n(n-7)+2(n-7)=0\\ (n+2)(n-7)=0\\ n=-2 \vee n=7$

Mind Peace answered Jun 12, 2015

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2 votes

(N-7)(n+2)=0 as you'd get n^2-7n+2n-14 then just simplify. Then it would be n=-2 or 7

Sherwin Zadeh answered Jun 17, 2015

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