(x+4)^2+(x-8)^2=2(8-x)(x+4) Решите пж

Решите задачу:

${(x + 4)}^{2} + {(x + 8)}^{2} = 2(8 - x)(x + 4) \\ {x}^{2} + 8x + 16 + {x}^{2} + 16x + 64 = 2(8x + 32 - {x}^{2} - 4x) \\ 2 {x}^{2} + 24x + 80 = 16x + 64 - 2 {x}^{2} - 8x \\ 2 {x}^{2} + 24x + 80 - 16x - 64 + 2 {x}^{2} + 8x = 0 \\ 4 {x}^{2} + 16x + 16 = 0( \div 4) \\ {x}^{2} + 4x + 4 = 0 \\ a = 1 \: \: b = 4 \: \: c = 4 \\ d = {b}^{2} - 4ac \\ d = {4}^{2} - 4 \times 1 \times 4 \\ d = 16 - 16 \\ d = 0 \\ x = \frac{ - 4}{2} \\ x = - 2$