Does any one know how to solve this -4n^2-2n=-6-7n-5n^2

2 answers:

6 0

$-4n^2-2n=-6-7n-5n^2 \\ -4n^2-2n+6+7n+5n^2=0\\n^2+5n+6=0 \\ n^2+2n+3n+6=0 \\ n(n+2)+3(n+2)=0 \\ (n+2)(n+3)=0 \\ n+2=0 \ \ \ \ or \ \ \ \ n+3=0\\n=-2 \ \ \ \ \ \ \ \ \ \ \ \ \ n=-3$

Answer: n=-2, n=-3

zaharov [31]3 years ago

3 0

-4n^2-2n=-6-7n-5n^2

Add 5n^2 to both sides
n^2-2n=-6-7n

Add 7n to both sides
n^2+5n=-6

Add 6 to both sides to put the equation into quadratic standard form
n^2+5n+6=0

Factor the equation
(n+2)(n+3)=0

3+2=5, 3*2=6

To solve a quadratic equation, find roots using Zero Product Property.
(n+2)=0
n=-2

(n+3)=0
n=-3

Final answers: n=-3, n=-2

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