For the quadratic equation $x^2-2\left(a+1\right)x+9a-5=0$, which of the following is/are true ?

The given quadratic equation is $x^2+2\left(a-1\right)x+9a-5=0$

Since the roots are real, the discriminant of the above equation is non negative

Hence, 

     $\begin{array}{rcl}{b}^2-4ac& =& 4{\left(a+1\right)}^2-4\left(9a-5\right)>0\\ & \Rightarrow & \left(a^2+2a+1\right)-9a+5>0\\ & \Rightarrow & a^2-7a+6>0\end{array}$

it implies that 

 Both roots are in opposite sign then its product is negative 

it implies that $9a-5<0\Rightarrow a<\frac{5}{9}$

If both roots are negative then the sum is negative

$2\left(a+1\right)>0\Rightarrow a>-1$