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The square of a number exceeds that number by 12. What are the two possible solutions?

3 or -4
3 or 4
-3 or 4

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Naǫid
x - the number

[tex]x^2=x+12 \\ x^2-x-12=0 \\ x^2+3x-4x-12=0 \\ x(x+3)-4(x+3)=0 \\ (x-4)(x+3)=0 \\ x-4=0 \ \lor \ x+3=0 \\ x=4 \ \lor \ x=-3[/tex]

The two possible solutions are -3 or 4.
Spolston222id

-3 or 4 hope this helps

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