Solve the following quadratic equations by factorization:

We have been given that,

$(x-5)(x-6)=\frac{25}{(24)^{2}}$

$x^{2}-11 x+30-\frac{25}{576}=0$

$x^{2}-11 x+\frac{17255}{576}=0$

$x^{2}-\frac{145}{24} x-\frac{119}{24} x+\frac{17255}{576}=0$

$x\left(x-\frac{145}{24}\right)-\frac{119}{24}\left(x-\frac{145}{24}\right)=0$

$\left(x-\frac{119}{24}\right)\left(x-\frac{145}{24}\right)=0$

Therefore,

$x-\frac{119}{24}=0$

$x=\frac{119}{24}$

or,

$x-\frac{145}{24}=0$

$x=\frac{145}{24}$

Hence, $x=\frac{119}{24}=4 \frac{23}{24}$ or $x=\frac{145}{24}=6 \frac{1}{24}$.