Sure, let's solve the given equation step-by-step.
We start with the quadratic equation:
[tex]\[ 2y^2 - 5y = 1 \][/tex]
To solve it, we first move all terms to one side to set the equation to zero:
[tex]\[ 2y^2 - 5y - 1 = 0 \][/tex]
This is now in the standard form of a quadratic equation [tex]\(ay^2 + by + c = 0\)[/tex], where [tex]\(a = 2\)[/tex], [tex]\(b = -5\)[/tex], and [tex]\(c = -1\)[/tex].
We can solve this quadratic equation using the quadratic formula:
[tex]\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Plugging in the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex], we get:
[tex]\[ y = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} \][/tex]
Simplifying the expression inside the square root:
[tex]\[ y = \frac{5 \pm \sqrt{25 + 8}}{4} \][/tex]
[tex]\[ y = \frac{5 \pm \sqrt{33}}{4} \][/tex]
So, the solutions are:
[tex]\[ y = \frac{5 - \sqrt{33}}{4} \][/tex]
[tex]\[ y = \frac{5 + \sqrt{33}}{4} \][/tex]
Therefore, the correct choice is:
A. The solution(s) is/are
[tex]\[ y = \frac{5 - \sqrt{33}}{4}, \frac{5 + \sqrt{33}}{4} \][/tex]
These are the exact solutions using radicals as needed.
