Posted by **jessica** on Sunday, December 7, 2008 at 11:41am.

having a hard time with a couple of problems please help

(3x+5)^2+8=0

also

6-9=12 divided by y+5 y-5 y^2-25

- algebra -
**bobpursley**, Sunday, December 7, 2008 at 12:10pm
On the first, two ways...

Let me do the easiest. Notice that the first term for real x will always be positive, and the second term is positive, the sum is zero. So there are no real x.

(3x+5)^2=-8

take the square root of each side..

3x+5=sqrt (-8)= 2i sqrt2

then solve for x. It will be a complex number.

Second

Multiply both sides by (y^2-25)

6(y-5)-9(y+5)=12

multiply out, gather terms and solve for y.

- algebra -
**jessica**, Sunday, December 7, 2008 at 12:39pm
Thank you very much

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