3y^2+8-(2y^2+12) Math Question?

Question

three y squared plus eight minus (in parenthesis) two y squared plus twelve(end parenthesis) I'm Not Sure How to work this problem...but the answers to the ways i found were -3x4y .......and y^2-4......[x = times] does anyone know the correct answer?

Answer 1

(3y^2 + 8) - (2y^2 + 12)

= 3y^2 + 8 - 2y^2 - 12

= (3y^2 - 2y^2) + (8 - 12)

= 1y^2 + (-4)

= y^2 - 4

The 2nd answer that you got is when you solved it the right way.....

Answer 2

The most simplified i got was (y+2)(y-2). This was my work:

3y^2+8-(2y^2+12) distribute negative outside of paranthesis to inside

3y^2+8-2y^2-12 combine like terms

y^2-4 find the square root

(y-2)(y+2)  

hope it helped a little =)

Answer 3

y^2-4

Try substitution to check.  Make y=5

y^2-4 would equal 21.

3*25+8-(50+12)= 83-62 = 21.

Check.  :)