Answer:
![x=(5+√(17))/(4)\,\,\,and\,\,\,x=(5-√(17))/(4)](https://img.qammunity.org/2021/formulas/mathematics/middle-school/dij3biku05d0tuhx4nhq78l7ompieqj8n8.png)
which agrees with the last two options in the list of possible answers
(mark both)
Explanation:
We start by re-writing the quadratic equation in standard form:
![2x^2-5x+1=0](https://img.qammunity.org/2021/formulas/mathematics/middle-school/lr61hqep6s2vjjcmq007a9c0aaiox0mv6r.png)
Which can be solved by using the quadratic formula for a quadratic
![(ax^2+bx+c=0)](https://img.qammunity.org/2021/formulas/mathematics/middle-school/7601ueoj23hsnsszx3w3cu703qan4lgwh6.png)
with parameters:
![a=2,\,\,b=-5,\,\,and\,\,c=1](https://img.qammunity.org/2021/formulas/mathematics/middle-school/9y6e2cut7erdxll95r1h7u95mlcd00b25y.png)
![x=(5+-√(25-4(2)(1)) )/(2*2) \\x=(5+-√(17))/(4)](https://img.qammunity.org/2021/formulas/mathematics/middle-school/xeatap8p2kv3xqyftt3a17mz02s8dcld1c.png)
Therefore the two values are:
![x=(5+√(17))/(4)\,\,\,and\,\,\,x=(5-√(17))/(4)](https://img.qammunity.org/2021/formulas/mathematics/middle-school/dij3biku05d0tuhx4nhq78l7ompieqj8n8.png)