Factor out $2$. Polynomial $6x^{2}-6x+7$ is not factored since it does not have any rational roots.
$$2\left(6x^{2}-6x+7\right)$$
Steps Using the Quadratic Formula
Quadratic polynomial can be factored using the transformation $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$, where $x_{1}$ and $x_{2}$ are the solutions of the quadratic equation $ax^{2}+bx+c=0$.
$$12x^{2}-12x+14=0$$
All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.
Since the square root of a negative number is not defined in the real field, there are no solutions. Quadratic polynomial cannot be factored.
$$12x^{2}-12x+14$$
Steps Using Direct Factoring Method
Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form $x^2+Bx+C=0$.This is achieved by dividing both sides of the equation by $12$
$$x ^ 2 -1x +\frac{7}{6} = 0$$
Let $r$ and $s$ be the factors for the quadratic equation such that $x^2+Bx+C=(x−r)(x−s)$ where sum of factors $(r+s)=−B$ and the product of factors $rs = C$
$$r + s = 1 $$ $$ rs = \frac{7}{6}$$
Two numbers $r$ and $s$ sum up to $1$ exactly when the average of the two numbers is $\frac{1}{2}*1 = \frac{1}{2}$. You can also see that the midpoint of $r$ and $s$ corresponds to the axis of symmetry of the parabola represented by the quadratic equation $y=x^2+Bx+C$. The values of $r$ and $s$ are equidistant from the center by an unknown quantity $u$. Express $r$ and $s$ with respect to variable $u$.
$$r = \frac{1}{2} - u$$ $$s = \frac{1}{2} + u$$
To solve for unknown quantity $u$, substitute these in the product equation $rs = \frac{7}{6}$
