Answer:
[tex]\frac{5}{4} +/-\frac{\sqrt{41} }{4}[/tex]
Step-by-step explanation:
We re-write the equation subtracting from both sides 3, so as to get a quadratic expression equal to zero and be able to use the quadratic formula to solve it:
[tex]2x^2-5x+1=3\\2x^2-5x+1-3=0\\2x^2-5x-2=0[/tex]
Now we use the quadratic formula considering that [tex]a=2, b=-5,\,and\,\,c=-2[/tex]
[tex]x=\frac{-b+/-\sqrt{b^2-4\,a\,c} }{2\,a} \\x=\frac{-(-5)+/-\sqrt{(-5)^2-4\,(2)\,(-2)} }{2\,(2)} \\\\x=\frac{5+/-\sqrt{25+16} }{4} \\x=\frac{5+/-\sqrt{41} }{4} \\x=\frac{5}{4} +/-\frac{\sqrt{41} }{4}[/tex]
