Answer
$x=1$ or $x=1+\frac{1}{a}$
Work Step by Step
$ax^2-(2a+1)x+(a+1)=0$
Solving for the trinomial using the quadratic formula for the quadratic equation of $ax^2+bx+c$, $x=\frac{-b \pm \sqrt {b^2-4ac}}{2a}$.
Thus, In this case,
$ax^2-(2a+1)x+(a+1)$, $a=a, b=-(2a+1), c=(a+1)$
Therefore,
$x=\frac{2a+1 \pm \sqrt {(2a+1)^2-4\times a(a+1)}}{2a}=\frac{2a+1 \pm \sqrt {4a^2+4a+1-4a^2-4a}}{2a}=\frac{2a+1\pm1}{2a}$
thus, $x=1$ or $x=1+\frac{1}{a}$
