Question 34247
*[invoke complete_square "x", 1, 6, -8]

*[invoke quadratic "z", 1, -6, -14]

{{{(x+4)(x-2)}}}<=0
{{{x+4=0}}}
{{{x=-4}}}
{{{x-2=0}}}
{{{x=2}}}
so answer is (- infinity,2]

Hope this helps,
lyra