Question 145099
Read this -->http://www.purplemath.com/modules/sqrquad.htm

Now let's solve the first one using that method. The others are for you

{{{x^2+8x+3=0}}}

Move the loose number over to the other side.
{{{x^2+8x=-3}}}

Divide through by whatever is multiplied on the squared term. 
In this case that is 1, so we are done.

Take half of the coefficient (don't forget the sign!) of the x-term, and square it. Add this square to both sides of the equation.
{{{8*(1/2)}}} = {{{4}}}
{{{4^2}}}={{{16}}}
{{{x^2+8x + 16=-3 + 16}}}

Convert the left-hand side to squared form, and simplify the right-hand side. (This is where you use that sign that you kept track of earlier.)
{{{(x+4)^2 =13}}}

Square-root both sides, remembering the "±" on the right-hand side.  Simplify as necessary.
{{{x+4 = +-sqrt(13)}}}

Solve for "x =".
{{{x = +-sqrt(13) - 4 }}}

{{{x = sqrt(13) - 4}}}  and {{{x = -sqrt(13) - 4}}}