First note that if [tex]m[/tex] is real-valued, then [tex]\sqrt{4m+49}[/tex] only exists if [tex]4m+49\ge0[/tex], or [tex]m\ge-\dfrac{49}4=-12.25[/tex].
Square both sides to get
[tex](m+1)^2=(\sqrt{4m+49})^2\implies m^2+2m+1=4m+49\implies m^2-2m-48=0[/tex]
This is easily factorized:
[tex]m^2-2m-48=(m-8)(m+6)\implies m=8\text{ or }m=-6[/tex]
Both of these solutions are larger than [tex]-\dfrac{49}4[/tex], so they are both valid solutions.
