Answer:
[tex]x=2.5+\frac{\sqrt{29}}{2}\\\\x=2.5-\frac{\sqrt{29}}{2}[/tex]
Step-by-step explanation:
So the question here is asking you to use the quadratic formula which is expressed as: [tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\[/tex]
A quadratic can generally be expressed as: [tex]y=ax^2+bx+c[/tex]
So using the equation you gave: [tex]0=x^2-5x-1[/tex]
We can identify the following values: a=1, b=-5, c=-1
Btw the equation explicitly write "1" as the coefficient of x, but since it's not provided it's implied that it's 1.
So plugging in the known values, we get the following equation:
[tex]x=\frac{5\pm\sqrt{(-5)^2-4(1)(-1)}}{2(1)}\\\\x=\frac{5\pm\sqrt{25+4}}{2}\\\\x=\frac{5\pm\sqrt{29}}{2}\\\\x=2.5+\frac{\sqrt{29}}{2}\\\\x=2.5-\frac{\sqrt{29}}{2}[/tex]
The last step just consists of taking the + and - solution, and since it asks for exact solutions you leave the 29 under the radical, and you don't approximate. There is no further simplification that can be done here.
