Answer:
[tex]x=\frac{5+\sqrt{17}}{4}\,\,\,and\,\,\,x=\frac{5-\sqrt{17}}{4}[/tex]
which agrees with the last two options in the list of possible answers
(mark both)
Step-by-step explanation:
We start by re-writing the quadratic equation in standard form:
[tex]2x^2-5x+1=0[/tex]
Which can be solved by using the quadratic formula for a quadratic [tex](ax^2+bx+c=0)[/tex]
with parameters:
[tex]a=2,\,\,b=-5,\,\,and\,\,c=1[/tex]
[tex]x=\frac{5+-\sqrt{25-4(2)(1)} }{2*2} \\x=\frac{5+-\sqrt{17}}{4}[/tex]
Therefore the two values are:
[tex]x=\frac{5+\sqrt{17}}{4}\,\,\,and\,\,\,x=\frac{5-\sqrt{17}}{4}[/tex]
