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