Question 597275
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -2x^2\ -\ 12x\ -\ y\ -\ 21\ =\ 0]


Step 1:  Move the *[tex \LARGE x^2] and *[tex \LARGE x] terms to the RHS:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ -y\ -\ 21\ =\ 2x^2\ +\ 12x]


Step 2:  In the RHS only, factor out the lead coefficient:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ -y\ -\ 21\ =\ 2(x^2\ +\ 6x)]


Step 3:  Divide the coefficient on the first degree *[tex \Large x] term by 2, square the result, and add that result <i><b>inside</b></i> the parentheses.  Multiply that value by the lead coefficient and add to the LHS to compensate:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ -y\ -\ 21\ +\ 18\ =\ 2(x^2\ +\ 6x\ +\ 9)]


Step 4:  Collect terms in the LHS and factor the perfect square trinomial in the RHS:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ -y\ -\ 3\ =\ 2(x\ +\ 3)^2]


Step 5:  Add 3 to both sides, then multiply by -1:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ y\ =\ -2(x\ +\ 3)^2\ -\ 3]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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