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].
