Answer
No real solution.
Work Step by Step
First transfer the constant to the right of the equal sign proceed with completing the square in line $1$ of the equation by dividing the coefficient of the $12m$ term by $2$ and squaring it and adding the result on both sides.
$$
\begin{aligned}
x^2+4x+7 & = 0 \\
x^2+4x & = -7 \\
x^2+4x+\left(\frac{4}{2}\right)^2 & =-7+\left(\frac{4}{2}\right)^2 \\
x^2+4x+4^2 & =-7+6^2 \\
(x+4)^2 & =-3.
\end{aligned}
$$ If the solution is to be a real number, then we can't take the square root of $-3$.
