Помогите решить,если можно с объяснением

1)
${(2x + 4)}^{2} = {(3x - 8)}^{2} \\ |2x + 4| = |3x - 8| \\ 2x + 4 = 3x - 8 \\ - x = - 12 \\ x = 12 \\ 2x + 4 = - 3x + 8 \\ 5x = 4 \\ x = \frac{4}{5}$
2)
${(6x + 2)}^{2} =( 3x - 2)^{2} \\ |6x + 2| = |3x - 2 | \\ 6x + 2 = 3x - 2 \\ 3x = - 4 \\ x = - \frac{4}{3} \\ 6x + 2 = - 3x + 2 \\ 9x = 0 \\ x = 0$
3)
$( {7x}^{2} + 8) {}^{2} = {(7x + 7)}^{2} \\ 7x {}^{2} + 8 = |7x + 7| \\ a)7x + 7 \geqslant 0 \\ x \geqslant - 1 \\ 7x {}^{2} + 8 = 7x + 7 \\ 7 {x}^{2} - 7x + 1 = 0 \\ d = 49 - 28 = 21 = \sqrt{21} \\ x = \frac{7 + \sqrt{21} }{14} \\ x = \frac{7 - \sqrt{21} }{14} \\ b)7x + 7 < 0 \\ x < - 1 \\7 {x}^{2} + 8 = - 7x - 7 \\ 7 {x}^{2} + 7x + 15 = 0 \\ d = 49 - 4 \times 7 \times 15 = 49 - 420 \\ d < 0$
4)
$(2 {x}^{2} + 2) {}^{2} = {x}^{4} \\ 2 {x}^{2} + 2 = {x}^{2} \\ {x}^{2} = - 2$
Пустое множество