2log_(1x^3)-8x+17vert 2:(3x^2+5)leqslant log_(3)3-8x+17:2x^2+7x+5}

To solve the given inequality, we need to simplify and analyze each part of the expression.<br /><br />First, let's simplify the expression inside the absolute value:<br /><br />$2\log_{1x^{3}} - 8x + 17$<br /><br />Since $\log_{1x^{3}}$ is not a valid logarithm (the base must be positive and not equal to 1), we cannot simplify this part further.<br /><br />Next, let's simplify the expression inside the parentheses:<br /><br />$(3x^{2} + 5) \leqslant \log_{3}3 - 8x + 17$<br /><br />Since $\log_{3}3 = 1$, we can rewrite the inequality as:<br /><br />$(3x^{2} + 5) \leqslant 1 - 8x + 17$<br /><br />Simplifying the right side, we get:<br /><br />$(3x^{2} + 5) \leqslant 18 - 8x$<br /><br />Rearranging the terms, we have:<br /><br />$3x^{2} + 8x - 13 \leqslant 0$<br /><br />Now, let's solve the quadratic inequality:<br /><br />$3x^{2} + 8x - 13 = 0$<br /><br />Using the quadratic formula, we find the roots of the equation:<br /><br />$x = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 3 \cdot (-13)}}{2 \cdot 3}$<br /><br />$x = \frac{-8 \pm \sqrt{64 + 156}}{6}$<br /><br />$x = \frac{-8 \pm \sqrt{220}}{6}$<br /><br />$x = \frac{-8 \pm 2\sqrt{55}}{6}$<br /><br />$x = \frac{-4 \pm \sqrt{55}}{3}$<br /><br />So, the solutions to the quadratic equation are:<br /><br />$x_1 = \frac{-4 + \sqrt{55}}{3}$ and $x_2 = \frac{-4 - \sqrt{55}}{3}$<br /><br />Now, let's consider the absolute value expression:<br /><br />$2\log_{1x^{3}} - 8x + 17 \geqslant 0$<br /><br />Since $\log_{1x^{3}}$ is not valid, we cannot solve this inequality.<br /><br />Therefore, the given inequality cannot be solved due to the invalid logarithmic expression.