Question 790486
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Completing the square:


Step 1:  Add the additive inverse of the constant term to both sides.


Step 2:  Multiply both sides by the reciprocal of the lead coefficient.


Step 3:  Divide the 1st degree term coeficient by 2, square the result, and add that result to both sides.  Combine terms in the LHS.


Step 4:  Factor the resulting perfect square trinomial in the RHS.


Step 5:  Take the square root of both sides, considering both positive and negative roots.


Step 6:  Add the additive inverse of the RHS constant to both sides.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x^2\ -\ 5x\ -\ 12\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x^2\ -\ 5x\ \ \ \ =\ 12]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ \frac{5}{2}x\ \ \ \ =\ 6]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ \frac{5}{2}x\ +\ \frac{25}{16}\ =\  \frac{121}{16}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(x\ -\ \frac{5}{4}\right)^2\ =\ \frac{121}{16}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ -\ \frac{5}{4}\ =\ \pm\ \frac{11}{4}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 4]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ -\frac{3}{2}]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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