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 $ax^{2}+bx+c=0$.
$$10p^{2}+3p-5=0$$
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 $p=\frac{-3±\sqrt{209}}{20}$ when $±$ is plus. Add $-3$ to $\sqrt{209}$.
$$p=\frac{\sqrt{209}-3}{20}$$
Now solve the equation $p=\frac{-3±\sqrt{209}}{20}$ when $±$ is minus. Subtract $\sqrt{209}$ from $-3$.
$$p=\frac{-\sqrt{209}-3}{20}$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $\frac{-3+\sqrt{209}}{20}$ for $x_{1}$ and $\frac{-3-\sqrt{209}}{20}$ 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 $10$
$$x ^ 2 +\frac{3}{10}x -\frac{1}{2} = 0$$
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 $rs = C$
$$r + s = -\frac{3}{10} $$ $$ rs = -\frac{1}{2}$$
Two numbers $r$ and $s$ sum up to $-\frac{3}{10}$ exactly when the average of the two numbers is $\frac{1}{2}*-\frac{3}{10} = -\frac{3}{20}$. 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 $u$.
