1) you can expand the brackets, simplify, set the equation = 0 and then do a double bracket factorisation.
[tex](x - 4)^{2} - 28 = 8 \\ x - 8x + 16 - 28 = 8 \\ x - 8x - 12 = 8 \\ x - 8x - 20 = 0 \\ (x - 10)(x + 2) = 0 \\ x = 10 \\ x = - 2[/tex]
2)
[tex]x^{2} - 6x - 7 = 0 \\ (x - 7)(x + 1) = 0 \\ \\ x - 7 = 0 \\ x = 7 \\ \\ x + 1 = 0 \\ x = - 1[/tex]
therefore Josh's solution is wrong asx = 7 and x= -1
