При каких значениях x значение дроби: (19-22x)/(13-2x) равно -5? (24x^2+37)/(2-x) равно...

Решите задачу:

$\frac{19-22x}{13-2x} = -5$
$19 - 22x = -5 * (13 - 2x)$
$19 - 22x = -65 + 10x$
$32x = 84$
$x = \frac{84}{32}$
$x = \frac{21}{8}$

$\frac{24 x ^ 2 + 37}{2 - x} = 17$
$24x ^ 2 + 37 = 17 * (2 - x)$
$24x ^ 2 + 37 = 34 - 17x$
$24x ^ 2 + 17x + 3 = 0$
$D = 17^2 - 4 * 3 * 24 = 289 - 288 = 1$
$x_{1} = \frac{- 17 + \sqrt{1} }{2 * 24} = \frac{-16}{48} = - \frac{1}{3}$
$x_{2} = \frac{- 17 - \sqrt{1} }{2 * 24} = \frac{-18}{48} = - \frac{3}{8}$