Answer:
[tex]x=\frac{5+\sqrt{41}}{4}[/tex] and [tex]x=\frac{5-\sqrt{41}}{4}[/tex]
Step-by-step explanation:
[tex]2x^2-5x+1=3[/tex]
To solve for x, we make right hand side 0
Subtract 3 from both sides
Given equation becomes
[tex]2x^2-5x-2=0[/tex]
We apply quadratic formula to solve for x
[tex]x=\frac{-b+-\sqrt{b^2-4ac}}{2a}[/tex]
a=2, b=-5, c=-2
[tex]x=\frac{5+-\sqrt{5^2-4(2)(-2)}}{2(2)}[/tex]
[tex]x=\frac{5+-\sqrt{41}}{4}[/tex]
[tex]x=\frac{5+\sqrt{41}}{4}[/tex] and [tex]x=\frac{5-\sqrt{41}}{4}[/tex]
