Question 410187
{{{z^2 + 8z + 7 <= 0}}}
Solve by completing the square
{{{z^2 + 8z + (8/2)^2 <= (8/2)^2 - 7}}}
{{{z^2 + 8z + 16 <= 16 - 7}}}
{{{(z + 4)^2 <= 11}}}
{{{z + 4 <= sqrt(11)}}}
{{{z <= sqrt(11) - 4}}}
and 
{{{z <= -sqrt(11) - 4}}}