Solve the equation $\left(\mathrm x-5\right)\left(\mathrm x-6\right)=\dfrac{25}{24^2}$ for x.

Hint: This equation can be solved easily be using the quadratic formula which is given by the formula-
$\dfrac{-\mathrm b\pm\sqrt{\mathrm b^2-4\mathrm{ac}}}{2\mathrm a}$
Here, a, b, c are constants and the equation is of the form ${ax}^{2}+{bx}+{c}=0$.

Complete step-by-step answer:
This equation can be solved by first converting it into general form and then applying quadratic formula.
$\left(\mathrm x-5\right)\left(\mathrm x-6\right)=\dfrac{25}{24^2}\\\mathrm{Let}\;\mathrm x-6\;=\;\mathrm p,\mathrm{then}\;\mathrm x-5=\mathrm p+1\\\mathrm{Let}\;\;\mathrm q=\dfrac5{24},\;\mathrm{then}\;\mathrm q^2=\dfrac{25}{24^2}\\\mathrm{The}\;\mathrm{equation}\;\mathrm{becomes}-\\\mathrm p\left(\mathrm p+1\right)=\mathrm q^2\\\mathrm p^2+\mathrm p-\mathrm q^2=0\\$
This is a quadratic equation in p. We can solve this simpler equation for p and then find the corresponding values of x.
Using the quadratic formula
$\dfrac{-\mathrm b\pm\sqrt{\mathrm b^2-4\mathrm{ac}}}{2\mathrm a}$
                             ,
$\Rightarrow \mathrm p=\dfrac{-1\pm\sqrt{1^2-4\times1\times\left(-\mathrm q^2\right)}}2\\\mathrm p=\dfrac{-1\pm\sqrt{1+4\mathrm q^2}}2\\\mathrm \Rightarrow p=\dfrac{-1\pm\sqrt{1+4\times{\displaystyle\dfrac{25}{576}}}}2\\\mathrm \Rightarrow p=\dfrac{-1\pm\sqrt{1+{\displaystyle\dfrac{25}{144}}}}2=\dfrac{-1\pm\sqrt{\displaystyle\dfrac{169}{144}}}2=\dfrac{-1\pm{\displaystyle\dfrac{13}{12}}}2\\\mathrm \Rightarrow p=\dfrac1{24},\;-\dfrac{25}{24}\\$
We assumed that p = x - 6, so the values for x will be-
x = p + 6
$\mathrm x=\;\dfrac1{24}+6=\dfrac{145}{24}\\\mathrm x=-\dfrac{25}{24}+6=\dfrac{119}{24}$
These are the required values of x.

Note: There is an alternative and shorter method to solve this problem. This can be done by splitting the term method.
Let x - 6 = a, so x - 5 = a + 1
Equation can be written as-
$\Rightarrow a\left(a+1\right)=\dfrac{25}{24\times24}\\a\left(a+1\right)=\dfrac1{24}\times\left(\dfrac{24+1}{24}\right)\\a\left(a+1\right)=\dfrac1{24}\left(1+\dfrac1{24}\right)\\$

$a^2+a-\dfrac1{24}\left(1+\dfrac1{24}\right)=0\\a^2-\left(1+\dfrac1{24}\right)a+\dfrac1{24}a-\dfrac1{24}\left(1+\dfrac1{24}\right)=0\\\left(a+\dfrac1{24}\right)\left(a-\dfrac{25}{24}\right)=0$
From these values of a find the corresponding values of x.