Помогите пожалуйста. Очень нужно..

$\frac{2}{a - 3} + 1 = \frac{15}{ {a}^{2} - 6a + 9}$
a≠3
$\frac{2}{a - 3} + 1 - \frac{15}{ {a}^{2} - 6a + 9} = 0$
$\frac{2}{a - 3} + 1 - \frac{15}{ {(a - 3)}^{2} } = 0$
$\frac{2(a - 3) + {(a - 3)}^{2} - 15}{ {(a - 3)}^{2} } = 0$
$\frac{2a - 21 + {a}^{2} - 6a + 9 }{ {(a - 3)}^{2} } = 0$
$\frac{ - 4a - 12 + {a}^{2} }{ {(a - 3)}^{2} } = 0$
$- 4a - 12a + {a}^{2} = 0$
${a}^{2} - 4a - 12 = 0$
$d = {( - 4)}^{2} - 4 \times 1 \times ( - 12) = 64$
$a1 = \frac{4 - 8}{2} = - 2$
$a2 = \frac{4 + 8}{2} = 6$
Ответ: (-2;6)