Answer:
[tex]\tt x_1=\cfrac{9-\sqrt{17}}{4}; \ \ x_2=\cfrac{9+\sqrt{17}}{4}[/tex]
Step-by-step explanation:
[tex]\displaystyle\tt 2x^2-9x+8=0\\\\x=\frac{-bб \sqrt{b^2-4ac}}{2a} \\\\x=\frac{9б\sqrt{(-9)^2-4\cdot2\cdot8}}{2\cdot2} \\\\x=\frac{9б\sqrt{81-64}}{4}\\\\x=\frac{9б\sqrt{17}}{4}\\\\x_1=\frac{9- \sqrt{17}}{4}\\\\x_2=\frac{9+ \sqrt{17}}{4}[/tex]
