For what values of m are the roots of $x^2 +2x+3 = m(2x+1)$ real and positive I am only able to show that to be real, $m <-1$ or $m\geq2$
Don't know how to finish solution
Answer is $2 \leq m < 3$
So far:
After expanding and factorising, 
$x^2 + 2(1-m)x + (3-m) = 0  $
Roots are real if discriminant $\ge 0 $ 
i.e. $4(m-2)(m+1)>=0  $
Therefore $m ≤ -1 \text{ or } m \ge 2$ for roots to be real
However I don't know how to find m for roots to be positive

$x^2 +2x+3 = m(2x+1)$
$x^2 +2(1-m)x+3-m = 0$
$D=b^2-4ac$
$D=4(1-m)^2-4(3-m)$
$D=4(1+m^2-2m)-12+4m$
$D=4+4m^2-8m)-12+4m$
$D=4m^2-4m-8$
For $D \geq 0$,you will find real roots
roots are real for  $m ≤ -1 \text{ or } m \ge 2$------(1)
$ x = (m-1) \pm \sqrt{m^2-m-2} $
roots are positive if $(m-1)>\sqrt{m^2-m-2}$
$m<3$---------------(2)
Combining (1) and (2) condition ,we get
roots are real and positive if $  2 \leq m<3$
