Answer:The solutions are [tex]x_{1}=2\ and\ x_{2}=0.5[/tex].Step-by-step explanation:Given:[tex]2x^{2} -5x+2=0[/tex]Now, to solve by using quadratic formula:[tex]x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}[/tex]Here [tex]a=2,b=-5\ and\ c=2[/tex][tex]x=\frac{-(-5)\pm \sqrt{(-5)^{2}-4\times 2\times 2}}{2\times 2}[/tex]Now, here we will take either [tex]x=\frac{-(-5)+\sqrt{25-16}}{4}[/tex] and then [tex]x=\frac{-(-5)-\sqrt{25-16}}{4}[/tex].[tex]x=\frac{-(-5)+\sqrt{25-16}}{4}[/tex] [tex]x=\frac{-(-5)-\sqrt{25-16}}{4}[/tex][tex]x=\frac{5+\sqrt{9}}{4}[/tex] [tex]x=\frac{5-\sqrt{9}}{4}[/tex][tex]x=\frac{5+3}{4}[/tex] [tex]x=\frac{5-3}{4}[/tex][tex]x=\frac{8}{4}[/tex] [tex]x=\frac{2}{4}[/tex][tex]x_{1}=2[/tex] [tex]x_{2}=\frac{1}{2}=0.5[/tex]Therefore, the solutions are [tex]x_{1}=2\ and\ x_{2}=0.5[/tex].