<PRE><B>Question 570028</B>
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&lt;i&gt;&lt;b&gt;Step 1.&lt;/b&gt;&lt;/i&gt;  Move the constant term to the right hand side.  This step is already done.


&lt;i&gt;&lt;b&gt;Step 2.&lt;/b&gt;&lt;/i&gt;  Divide through by the lead coefficient.  The lead coefficient in this problem is 1, so you can skip this step.


&lt;i&gt;&lt;b&gt;Step 3.&lt;/b&gt;&lt;/i&gt;  Divide the coefficient on the first degree term by 2.  In this case, 2 divided by 2 is 1.


&lt;i&gt;&lt;b&gt;Step 4.&lt;/b&gt;&lt;/i&gt; Square the result of Step 3.  1 squared is 1.


&lt;i&gt;&lt;b&gt;Step 5.&lt;/b&gt;&lt;/i&gt; Add the result of Step 4 to both sides of the equation.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 2x\ +\ 1\ =\ 49]


&lt;i&gt;&lt;b&gt;Step 6.&lt;/b&gt;&lt;/i&gt; Factor the perfect square trinomial in the LHS:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ + 1)^2\ =\ 49]


&lt;i&gt;&lt;b&gt;Step 7.&lt;/b&gt;&lt;/i&gt; Take the square root of both sides, considering both the positive and negative root:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ 1\ =\ \pm\sqrt{49}\ =\ \pm7]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ 1\ =\ 7]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ 1\ =\ -7]


&lt;i&gt;&lt;b&gt;Step 8.&lt;/b&gt;&lt;/i&gt; Solve the two linear equations to get your two roots.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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