<PRE><B>Question 201687</B>
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&lt;b&gt;Step 1:&lt;/b&gt;.  Divide both sides by the lead coefficient.  For your problem, the lead coefficient is 1, so there is nothing to do.


&lt;b&gt;Step 2:&lt;/b&gt;.  Add the opposite of the constant term on the left to both sides, i.e., combine the constant terms and put them on the right hand side:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ b^2 - 8b\ \ \ \ =\ 48]


&lt;b&gt;Step 3:&lt;/b&gt;.  Divide the coefficient on the 1st degree term by 2 and square the result, adding that value to both sides of the equation.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\frac{-8}{2}\right)^2 = 16], so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ b^2 - 8b + 16\ =\ 64]


&lt;b&gt;Step 4:&lt;/b&gt;.  Factor the perfect square trinomial on the left:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (b - 4)^2 = 64]


&lt;b&gt;Step 5:&lt;/b&gt;.  Take the square root of both sides, remembering to consider both the positive and negative roots.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ b - 4 = \pm \sqrt{64} = \pm 8]


&lt;b&gt;Step 6:&lt;/b&gt;.  Solve


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ b = 4 \pm 8]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ b = 12], or *[tex \LARGE b = -4]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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