How do you solve \[6{x^2} - 2 = x\]?

Hint: Here given a algebraic equation firstly we have to convert the given equation to the quadratic equation \[a{x^2} + bx + c = 0\] by shifting x from RHS to LHS. Later solve the quadratic equation by using the method of factorisation and find the factors and equate each factor to the zero to get the value of x.

Complete step by step solution:
Factorization means the process of creating a list of factors otherwise in mathematics, factorization or factoring is the decomposition of an object into a product of other objects, or factors, which when multiplied together give the original.

The general form of an equation is \[a{x^2} + bx + c\] when the coefficient of \[{x^2}\] is unity. Every quadratic equation can be expressed as \[{x^2} + bx + c = \left( {x + d} \right)\left( {x + e} \right)\]. Here, b is the sum of d and e and c is the product of d and e.
Consider the given expression:
\[ \Rightarrow \,\,\,6{x^2} - 2 = x\]
Subtract both side by x, then
\[ \Rightarrow \,\,\,6{x^2} - 2 - x = x - x\]
On simplification, we get
\[ \Rightarrow \,\,\,6{x^2} - x - 2 = 0\]
The above equation similar like a quadratic equation \[a{x^2} + bx + c\] now, solve by the method of factorization.
Now, Break the middle term as the summation of two numbers such that its product is equal to -12. Calculated above such two numbers are -4 and 3.
\[ \Rightarrow \,\,\,6{x^2} + 3x - 4x - 2 = 0\]
Making pairs of terms in the above expression
\[ \Rightarrow \,\,\,\left( {6{x^2} + 3x} \right) - \left( {4x + 2} \right) = 0\]
Take out greatest common divisor GCD from the both pairs, then
\[ \Rightarrow \,\,\,3x\left( {2x + 1} \right) - 2\left( {2x + 1} \right) = 0\]
Take \[\left( {2x + 1} \right)\] common
\[ \Rightarrow \,\,\,\left( {2x + 1} \right)\left( {3x - 2} \right) = 0\]
Equate the each factor to zero, then
\[ \Rightarrow \,\,\,\left( {2x + 1} \right) = 0\] or \[\left( {3x - 2} \right) = 0\]
\[ \Rightarrow \,\,\,2x = - 1\] \[3x = 2\]
\[ \Rightarrow \,\,\,x = - \dfrac{1}{2}\] \[x = \dfrac{2}{3}\]

Hence, the required solution is \[x = - \dfrac{1}{2}\] or \[x = \dfrac{2}{3}\].

Note: The equation is a quadratic equation. This problem can be solved by using the sum product rule. This defines as for the general quadratic equation \[a{x^2} + bx + c\], the product of \[a{x^2}\] and c is equal to the sum of bx of the equation. Hence we obtain the factors. The factors for the equation depend on the degree of the equation.