Решите легкое уравнение со степенями

$2 \times {4}^{3 {x}^{2} + 5x } - 4 = 7 \times {8}^{ {x}^{2} + \frac{5x}{3} } \\ \\ 2 \times {4}^{3 {x}^{2} + 5x } - 4 - 7 \times {2}^{3 {x}^{2} + 5x } = 0 \\$

${2}^{3 {x}^{2} + 5x} = t \\$

t > 0

$2 {t}^{2} - 4 - 7t = 0 \\ 2 {t}^{2} - 7t - 4 = 0 \\$

D = 49 - 4•2•(-4) = 49 + 32 = 81

$t1 = \frac{ 7 - 9}{4} = \frac{ - 2}{4} = - \frac{1}{2} \\ \\ t2 = \frac{7 + 9}{4} = \frac{16}{4} = 4 \\$

t = - 1/2 - не подходит, t > 0

Обратная замена:

$t = 4 \\ \\ {2}^{3 {x}^{2} + 5x } = {2}^{2} \\ \\ 3 {x}^{2} + 5x - 2 = 0 \\$

D = 25 - 4•3•(-2) = 25 + 24 = 49

$x1 = \frac{ - 5 - 7}{6} = \frac{ - 12}{6} = - 2 \\ \\ x2 = \frac{ - 5 + 7}{6} = \frac{2}{6} = \frac{1}{3} \\ \\$

ОТВЕТ: - 2 ; 1/3