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Mathematics 18 Online
OpenStudy (anonymous):

(2x + 2) (5x - 5) = 0

OpenStudy (luigi0210):

Wait, solving or distributing?

OpenStudy (anonymous):

its solving quadratic equations

OpenStudy (luigi0210):

Set each equal to zero then. That'll give you two x-values.

OpenStudy (anonymous):

what?

OpenStudy (anonymous):

Plug in each number starting from 0 and working up into x until the result is 0. That how I get it done fast.

OpenStudy (anonymous):

(2x + 2) (5x + 5) = 0 10x^2 + 10x + 10x + 10 = 0 10x^2 + 20x = -10 there someone else can help from there /)_-

OpenStudy (anonymous):

could you do this for an example because i dont get what your saying

OpenStudy (anonymous):

Ok so you have, (2x + 2) (5x + 5) = 0 Lets say x=0 and plug it in. (2(0)+2)*(5(0)+5)=? Do that and you get, 10. 10≠0 So change x=0, to x=1 and plug it in: (2(1)+2)*(5(1)+5). Do that and you'll find out that it results in 0. 0=0 so, x=1. @vausvega

OpenStudy (anonymous):

okay thank you

OpenStudy (anonymous):

z^2 - 6z - 27 = 0

OpenStudy (anonymous):

XD just do the same, start by replacing z with (0) or 0, and go up from there until the left side equals the right. (If the answer is above 10 or 20 you might need a better way of doing it.)

OpenStudy (anonymous):

the answer choices say: a.) z=3 or z-9 b.) z = 3 or z = -9 c.) z = 3 or z + -2 d.) z = -3 or z = -9

OpenStudy (luigi0210):

Guys.. if you're solving just do this ._. \(2x+2=0\) \(5x-5=0\) And solve..

OpenStudy (luigi0210):

No need to multiply it back out >_>

OpenStudy (anonymous):

O.o yea try ^his solution with this kind. I tried and couldn't figure.

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