Помогите пожалуйста (2x) / (x^2-3x+4)- (3x) / (x^2-6x+4)=4

ОДЗ: x²-3x+4≠0 и x²-6x+4≠0 ⇒х≠3-√5; х≠3+√5
$\frac{2x}{ x^{2} -3x+4} - \frac{3x}{ x^{2} -6x+4}=4\\ \frac{2x}{ x^{2} -3x+4} - \frac{3x}{ x^{2} -3x+4-3x}=4 \\ x^{2} -3x+4 =t \\ \frac{2x}{t} - \frac{3x}{ t-3x}=4 \\ 2x(t-3x)-3xt-4t(t-3)=0 \\ 2xt-6 x^{2} -3xt-4t^{2}+12xt=0 \\ 6x ^{2} -11tx+4 t^{2}= 0 \\ D=121 t^{2} -96 t^{2} =25 t^{2} ; \sqrt{D} = \sqrt{25 t^{2} } =5t \\ x_{1} = \frac{11t-5t}{12} = \frac{t}{2} \\ x_{2} = \frac{11t+5t}{12} = \frac{4t}{3} \\ 6x ^{2} -11tx+4 t^{2}= 6(x- \frac{t}{2} )(x- \frac{4t}{3})=0 \\$ $6(x- \frac{t}{2} )(x- \frac{4t}{3})=0 \\ (x- \frac{t}{2} )=0 \\ x- \frac{ x^{2} -3x+4}{2} =0 \\ 2x- x^{2} +3x-4=0 \\ - x^{2} +5x-4=0 \\ x^{2} -5x+4=0 \\ x_{1}=4;x_{2}=1 \\ x- \frac{4t}{3} =0 \\x- \frac{4( x^{2} -3x+4)}{3} =0 \\ 3x-4 x^{2} +12x-16=0 \\ -4 x^{2} +15x-16=0 \\ D=225-256\ \textless \ 0$
отв:1;4