Question 882939
Solve: {{{x^2-2x-15<0}}}
<pre>
{{{x^2 - 2x - 15 < 0}}}
{{{x^2 - 2x - 15 = 0}}} ---- Changing inequality to equality to determine CRITICAL POINTS.
(x + 3)(x - 5) = 0
Critical points: {{{x = - 3_AND_x = 5}}}
Test intervals:  x < - 3, {{{- 3 < x < 5}}}, and x > 5
ONLY {{{highlight_green(highlight_green(- 3 < x < 5))}}} proves TRUE.