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