Алгебра: Помогите решить уравнение!

По свойству:

$\sqrt[n]{ {a}^{m} } = {a}^{ \frac{m}{n} } \\ \\$

${ ({4}^{x + 2} )}^{x} \times \sqrt[3]{ {32}^{x - 1} } = 64 \\ \\ {2}^{2x(x + 2)} \times {2}^{ \frac{5(x - 1)}{3} } = {2}^{6} \\ \\ {2}^{2 {x}^{2} + 4x + \frac{5x - 5}{3} } = {2}^{6} \\ \\ 2 {x}^{2} + 4x + \frac{5x - 5}{3} = 6 \\ \\ 6 {x}^{2} + 12x + 5x - 5 = 18 \\ \\ 6 {x}^{2} + 17x - 23 = 0 \\ \\ d = {17}^{2} - 4 \times 6 \times ( - 23) = 841 \\ \\ 1) \: \: \: \: x = \frac{ - 17 - 29}{12} = \frac{ - 46}{12} = - \frac{23}{6} \\ \\ 2) \: \: \: \: x = \frac{ - 17 + 29}{12} = - \frac{12}{12} = - 1 \\ \\$

ОТВЕТ: - 23/6 ; - 1