Question

How do you solve this inequality? -x^2-3x+17_>0​

Answer 1

$\bf -x^2-3x+14\ge -3 \implies 0\ge x^2+3x-17 \\\\[-0.35em] ~\dotfill\\\\ ~~~~~~~~~~~~\textit{quadratic formula} \\\\ y=\stackrel{\stackrel{a}{\downarrow }}{1}x^2\stackrel{\stackrel{b}{\downarrow }}{+3}x\stackrel{\stackrel{c}{\downarrow }}{-17} \qquad \qquad x= \cfrac{ - b \pm \sqrt { b^2 -4 a c}}{2 a}$

$\bf x=\cfrac{-3\pm\sqrt{3^2-4(1)(-17)}}{2(1)}\implies x=\cfrac{-3\pm \sqrt{9+68}}{2} \\\\\\ x=\cfrac{-3\pm\sqrt{77}}{2} \\\\[-0.35em] ~\dotfill\\\\ 0\ge \left( x+\cfrac{-3+\sqrt{77}}{2} \right)\left( x+\cfrac{-3-\sqrt{77}}{2} \right) \\\\\\ \cfrac{3-\sqrt{77}}{2}\ge x\qquad and\qquad \cfrac{3+\sqrt{77}}{2}\ge x$