Question 1117503
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The way you wrote your equation is very odd, and I suspect you left out a plus or minus sign, probably between the closing bracket and the 6y in the RHS.  But I can only go with what you presented, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4(y\ +\ 1)\ -\ 7\ =\ -2(-8y\ +\ 7)6y]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4y\ +\ 4\ -\ 7\ =\ -12y(-8y\ +\ 7)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4y\ -\ 3\ =\ 96y^2\ -\ 84y]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 96y^2\ -\ 88y\ +\ 3\ =\ 0]


Using the quadratic formula:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ \frac{88\ \pm\ \sqrt{(-88)^2\ -\ 4(96)(3)}}{2(96)}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ \frac{11\ \pm\ \sqrt{103}}{24}]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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