Answer:
[tex]x_{1}=\frac{-3+\sqrt{13} }{2}\\[/tex]
[tex]x_{2}=\frac{-3-\sqrt{13} }{2}\\[/tex]
Step-by-step explanation:
the equation is: [tex]x^2-3x-1=0[/tex]
you have a quadratic equation of the form:
[tex]ax^2+bx+c=0[/tex]
where [tex]a=1,b=-3,c=-1[/tex]
this can be solved using the general formula:
[tex]x=\frac{-b+-\sqrt{b^2-4ac} }{2a}[/tex]
substituting all the known values:
[tex]x=\frac{-(-3)+-\sqrt{(-3)^2-4(1)(-1)} }{2(1)} \\x=\frac{-3+-\sqrt{9+4} }{2}\\ x=\frac{-3+-\sqrt{13} }{2}\\[/tex]
one value for x will be found using the plus sign before the square root:
[tex]x_{1}=\frac{-3+\sqrt{13} }{2}\\[/tex]
and the other will be found using the negative sign before the square root:
[tex]x_{2}=\frac{-3-\sqrt{13} }{2}\\[/tex]
