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 $ax^{2}+bx+c=0$.
$$10y^{2}-28y+14=0$$
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.
Now solve the equation $y=\frac{28±4\sqrt{14}}{20}$ when $±$ is plus. Add $28$ to $4\sqrt{14}$.
$$y=\frac{4\sqrt{14}+28}{20}$$
Divide $28+4\sqrt{14}$ by $20$.
$$y=\frac{\sqrt{14}+7}{5}$$
Now solve the equation $y=\frac{28±4\sqrt{14}}{20}$ when $±$ is minus. Subtract $4\sqrt{14}$ from $28$.
$$y=\frac{28-4\sqrt{14}}{20}$$
Divide $28-4\sqrt{14}$ by $20$.
$$y=\frac{7-\sqrt{14}}{5}$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $\frac{7+\sqrt{14}}{5}$ for $x_{1}$ and $\frac{7-\sqrt{14}}{5}$ for $x_{2}$.
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 $10$
$$x ^ 2 -\frac{14}{5}x +\frac{7}{5} = 0$$
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 $rs = C$
$$r + s = \frac{14}{5} $$ $$ rs = \frac{7}{5}$$
Two numbers $r$ and $s$ sum up to $\frac{14}{5}$ exactly when the average of the two numbers is $\frac{1}{2}*\frac{14}{5} = \frac{7}{5}$. 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 $u$.
$$r = \frac{7}{5} - u$$ $$s = \frac{7}{5} + u$$
To solve for unknown quantity $u$, substitute these in the product equation $rs = \frac{7}{5}$
