For what values of m does the quadratic equation (m-7)x2 + (2m-1)x + (2m+3) = 0 have real roots?

The given equation is $(m-7)x^2 + (2m-1)x + (2m+3) = 0$.
This is in the form $Ax^2 + Bx + C = 0$, where:
$A = m-7$
$B = 2m-1$
$C = 2m+3$

For a quadratic equation to have real roots, the discriminant $\Delta = B^2 - 4AC$ must be greater than or equal to zero.

Expand the terms:

Substitute these back into the inequality:

Combine like terms:

Multiply the entire inequality by $-1$ and reverse the inequality sign:

Using the quadratic formula $m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

Calculate the value of $\sqrt{2960}$:

Now find the two roots:

Since the quadratic $4m^2 - 40m - 85$ opens upwards (coefficient of $m^2$ is positive), the inequality $4m^2 - 40m - 85 \le 0$ holds for values of $m$ between its roots.
So, $-1.8007 \le m \le 11.8007$.

We also need to consider the case where $A = m-7 = 0$, which means $m=7$. If $m=7$, the original equation becomes $13x + 17 = 0$, which is a linear equation with one real root $x = -\frac{17}{13}$. Since $m=7$ results in a real root, it is included in the range. The value $m=7$ falls within the calculated interval.

$m_1 \approx -1.8$
$m_2 \approx 11.8$

The range of values for $m$ is $\boxed{-1.8 \le m \le 11.8}$.