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A good start is subtracting 64 from both sides.\[ \color{Black}{\Rightarrow 9x^2 + 3x + 25 - 64 = 0}\] Simplify, and solve the quadratic equation.
the 9x^2 is what I am confused about.
confused ??

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Other answers:

u can solve it by factorization...
9x2+30x+25=64 9x2+30x+25-64 = 0 9x2+30x-39 = 0 9x2 -9x + 39x -39 = 0 9x(x-1) +39 (x-1) = 0 (9x +39) (x-1) = 0 so, x = -39/9 and x =1....
sorry, but I do know the answer involves imaginary numbers.
Actually no... The solutions for x as an answer does not involve imaginary #'s, and it's not 59/3
@Ganpat is correct. my answer was wrong...sorry
With the imaginary line, it's the intersection: |dw:1343664593480:dw|
The answer in the back of the book is \[x=(-10+2i \sqrt14)/6, (-10-2i \sqrt14)/6 \] I just can't figure out how it got there.
The blue line is (3x+5)^2 The imaginary line is 64i
That answer is crazy, this has solvable real number values. Yes you can combine those to get X, but why?
In this case I'd say either the problem was written down for us incorrectly/incomplete-directions, or the book is making it a whole lot harder than it needs be...
i do not know.
:-S Go with what @Ganpat said, unless there's something missing or incorrect here with regards to the question and it's directions as you wrote it here. He's 100% correct.
Plug in those values for x and you'll it works.
@agentx5 : when did i say that ? :D.. lol
Huh? o_O
kidding dude !! :)

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