No, and there are a few ways to check his solution.
1. If [tex]x=-13[/tex], we get [tex]x^2-6x-7=240\neq0[/tex]. So -13 can't be a solution, and the same goes for 19.
2. The quadratic can be factorized pretty easily: [tex]x^2-6x-7=(x-7)(x+1)[/tex], which means the solutions should be [tex]x=7[/tex] and [tex]x=-1[/tex].
3. Check Josh's reasoning. The mistake occurs between the 4th and 5th/7th lines, where Josh wrote
[tex](x-3)^2=16\implies x-3=16[/tex]
[tex](x-3)^2=16\implies x-3=-16[/tex]
This is not true. He was supposed to take the square root of 16 first:
[tex](x-3)^2=16\implies x-3=\sqrt{16}=4\implies x=7[/tex]
[tex](x-3)^2=16\implies x-3=-\sqrt{16}=-4\implies x=-1[/tex]
