Solve the following quadratic equations by factorization:

We have been given

$\frac{x+3}{x+2}=\frac{3 x-7}{2 x-3}$

$2 x^{2}-3 x+6 x-9=3 x^{2}-7 x+6 x-14$

$x^{2}-4 x-5=0$

$x^{2}-5 x+x-5=0$

$x(x-5)+1(x-5)=0$

$(x+1)(x-5)=0$

Therefore,

$x+1=0$

$x=-1$

or,

$x-5=0$

$x=5$

Hence, $x=-1$ or $x=5$.