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(2x-7)^2-4(2x-7)-12=0, i see there are 2 (2x-7) in this equation. Is there a purpose for that? How do i go about this one, now??? thank you
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it might help to replace the (2x-7) with another variable, say z let z = 2x-7 so that allows us to go from this 2x-7)^2-4(2x-7)-12=0 to this z^2-4z-12=0
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solve z^2-4z-12=0 for z. You'll get 2 solutions. Each solution in terms of z will then be used to find the solutions in terms of x
how do i get 3 z's in a factoring.
to factor z^2-4z-12, you need to think of 2 numbers that multiply to -12 (last term) AND add to -4 (middle coefficient)
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what two numbers do this?
wait nevermind, i got it (u+2)(u-6)=0
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