Question 436855
<pre>
{{{4p^2-32p+29=0}}}

Get the constant term +29 off the left side:

{{{4p^2-32p=-29}}}


Divide through by the leading coefficient 4

{{{(4p^2)/4-(32p)/4=-29/4}}}

{{{p^2-8p=-29/4}}}

To the side, multiply the coefficient of p, which is -8
by 1/2, getting -4, then square -4, getting (-4)² = 16.
Then add +16 to both sides:

{{{p^2-8p+16=-29/4+16}}}

This causes the left side to became a trinomial which
when factored becomes a perfect square.  And we get an
LCD on the right of 4 


{{{(p-4)(p-4) = -29/4+64/4}}}

Write the left side as a perfect square.  Combine
the fractions on the right:

{{{(p-4)^2=35/4}}}

Use the principle of square roots:

{{{p-4= "" +- sqrt(35/4)}}}

We can take the square root of 4 on the bottom
on the right:

{{{p-4= "" +- sqrt(35)/2}}}

Solve for p by adding 4 to both sides:

{{{p=4 +- sqrt(35)/2}}}

You can leave it like that or you can
get an LCD of 2

{{{p=8/2 +- sqrt(35)/2}}}

And then combine the two fractions

{{{p=(8+- sqrt(35))/2}}}

The two solutions are:

{{{p=(8+ sqrt(35))/2}}} and {{{p=(8- sqrt(35))/2}}}

Edwin</pre>