Algebra and Trigonometry 10th Edition

{$\frac{-5-\sqrt {89}}{4}, \frac{-5+\sqrt {89}}{4}$}
$2x^2+5x-8=0$ $x^2+\frac{5x}{2}=4$ $x^2+\frac{5x}{2}+\frac{25}{16}=\frac{89}{16}$ $(x+\frac{5}{4})^2=\frac{89}{16}$ $(x+\frac{5}{4})=+\sqrt {\frac{89}{16}} or -\sqrt {\frac{89}{16}}=+{\frac{\sqrt{89}}{4}}or -{\frac{\sqrt{89}}{4}}$ x={$\frac{-5-\sqrt {89}}{4}, \frac{-5+\sqrt {89}}{4}$}