How do you solve ${{x}^{2}}-15x+54=0?$

Hint: As the general expression of any quadratic equation is $a{x^2} + bx + c = 0,a \ne 0$, where a,b and c are the value of constant.
Here, we select any two numbers such that whose product is equal to the product of ‘a’ and 'c’ and which when added or subtracted results in the value of “b”.
Like, ${x^2} + x - 2 = 0$ here, a = 1, b = 1 and c = - 2
So, product of “a” and “c” will be -2
So, two such numbers will be 2 and -1 which one multiplied result to -2 i.e. value of “c” and which one subtracted result to the value of “b” i.e. 1
Apply this concept to solve the equation given in the question.

Complete step-by-step solution:
We have given equation,
${{x}^{2}}-15x+54=0$
So, here $\left( x-9 \right)$ and $\left( x-6 \right)$ are the two linear factor of quadratic polynomial of ${{x}^{2}}-15x+54=0$
So, the quadratic equation obtained from ${{x}^{2}}-15x+54=0$ can be written as,
${{x}^{2}}-9x-6x+54=0$
$\Rightarrow x\left( x-9 \right)-6\left( x-9 \right)=0$
$\Rightarrow \left( x-6 \right)\left( x-9 \right)=0$
$\Rightarrow x-6=0$ or $x-9=0$
$\Rightarrow x-6$ or $x=9$

Here, $6$ and $9$ are roots of a given quadratic equation.

Note: $\left( 1 \right)$ Factoring in a polynomial has writing it as a product of two or more polynomials.
$\left( 2 \right)$ We have several examples of factoring. However for this you should take common factors using distributive property.
For example: $6{{x}^{2}}+4x=2x\left( 3x+2 \right)$
$\left( 3 \right)$ In the method of factorisation we reduce it to a simple fan. The factor can be an integer, a variable or algebraic itself in any equation.
$\left( 4 \right)$ There are various methods of factorisation; they are factoring out the GCF, the grouping method, the difference of square pattern etc.