Question 189423
{{{x^2 + 7x + 4 = 0}}}
Subtract {{{4}}} from both sides
{{{x^2 + 7x = -4}}}
Complete the square by taking {{{1/2}}} of the
coefficient of the {{{x}}} term, square it, and
add it to both sides
{{{x^2 + 7x + (7/2)^2 = -4 + (7/2)^2}}}
{{{x^2 + 7x + 49/4 = -4 + 49/4}}} 
{{{x^2 + 7x + 49/4 = -(16/4) + 49/4}}}
{{{x^2 + 7x + 49/4 = 33/4}}}
The left side is a perfect square. Rewrite it.
{{{(x + 7/2)^2 = 33/4}}}
Take the square root of both sides
{{{x + 7/2 = sqrt(33)/2}}}
and also
{{{x + 7/2 = -sqrt(33)/2}}}
so
{{{x = (-7 + sqrt(33))/2}}}
and
{{{x = (-7 - sqrt(33))/2}}}