<PRE><B>Question 1127312</B>
&lt;br&gt;
You are likely to get fractions when completing the square, except in &quot;nice&quot; problems....&lt;br&gt;
{{{8x^2+10x = 3}}}&lt;br&gt;
Factor out the leading coefficient:&lt;br&gt;
{{{8(x^2+(5/4)x) = 3}}}&lt;br&gt;
Take half the coefficient of the x term (5/8), square it (25/64), and add 25/64 in the parentheses on the left, to complete the square.  Note that you added 8*25/64 = 25/8 on the left side of the equation, so you need to add 25/8 on the right to keep the equation balanced.&lt;br&gt;
(Note that the only time you will NOT get fractions in the process of completing the square is when half the coefficient of x is an integer.)&lt;br&gt;
{{{8(x^2+(5/4)x+25/64) = 3+25/8}}}&lt;br&gt;
Write the expression in parentheses as the square of a binomial, and simplify the expression on the right:&lt;br&gt;
{{{8(x+(5/8))^2 = 49/8}}}&lt;br&gt;
Divide by 8 to get the squared binomial by itself:&lt;br&gt;
{{{x+(5/8)^2 = 49/64}}}&lt;br&gt;
Take the square root of both sides, getting both positive and negative roots on the right:&lt;br&gt;
{{{x+5/8 = 7/8}}}  OR  {{{x+5/8 = -7/8}}}&lt;br&gt;
Solve the two equations to find the two roots of the original equation:&lt;br&gt;
{{{x = 2/8 = 1/4}}}  OR  {{{x = -12/8 = -3/2}}}&lt;br&gt;
The two roots of the equation are 1/4 and -3/2.</PRE>