Question 235794
{{{ x^2+5x<=3 }}}
Solve it as equal, then add the <
{{{ x^2+5x<=3 }}}
{{{ x^2+5x-3=0 }}}
*[invoke solve_quadratic_equation 1,5,-3]
Sub <= for the equal signs.
All values between the 2 solutions fit the inequality.
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It's not -3 and 1.
For the solution to be -3 and 1:
{{{x^2 + 2x <=3}}}
--> -3>=x>=1