Solve the following inequality algebraically. -6x^2+15geqslant x+8 Answer Attemptiout of 2 square

1. Rearrange the inequality<br /> Move all terms to one side: $-6x^2 - x + 15 - 8 \geqslant 0$ simplifies to $-6x^2 - x + 7 \geqslant 0$.<br />2. Solve the quadratic equation<br /> Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ for $-6x^2 - x + 7 = 0$. Here, $a = -6$, $b = -1$, $c = 7$.<br /> Calculate discriminant: $b^2 - 4ac = (-1)^2 - 4(-6)(7) = 1 + 168 = 169$.<br /> Roots are $x = \frac{-(-1) \pm \sqrt{169}}{2(-6)} = \frac{1 \pm 13}{-12}$.<br /> Solutions: $x = \frac{14}{-12} = -\frac{7}{6}$ and $x = \frac{-12}{-12} = 1$.<br />3. Test intervals<br /> Test intervals $(-\infty, -\frac{7}{6})$, $(-\frac{7}{6}, 1)$, $(1, \infty)$ in $-6x^2 - x + 7 \geqslant 0$.<br /> Choose test points like $x = -2$, $x = 0$, $x = 2$.<br /> For $x = -2$: $-6(-2)^2 - (-2) + 7 = -24 + 2 + 7 = -15 < 0$ (False).<br /> For $x = 0$: $-6(0)^2 - 0 + 7 = 7 > 0$ (True).<br /> For $x = 2$: $-6(2)^2 - 2 + 7 = -24 - 2 + 7 = -19 < 0$ (False).<br />4. Determine solution set<br /> Inequality holds for $x \in [-\frac{7}{6}, 1]$.