Алгебра,Буду благодарна Всем добра♥

$\frac{15}{ x^{2}-6x+9 } - \frac{2}{x-3} = 1 \\ \frac{15}{ (x-3)^{2} } - \frac{2}{x-3} = 1 \\ \frac{15}{ (x-3)^{2} } - \frac{2(x-3)}{(x-3)(x-3)} = 1 \\$
$\frac{15-2(x-3)}{ (x-3)^{2} } = 1 \\ \frac{15-2x+6}{ (x-3)^{2} } = 1 \\$
$\frac{21-2x}{ (x-3)^{2} } = 1 \\ (x-3)^{2} = 21-2x \\ x^{2} -6x+9 + 2x - 21 = 0 \\$
$x^{2} - 4x - 12 = 0 \\ x_{1} = 6, x_{2} = -2 \\$

Ответ: -2 ; 6.