Factor the expression by grouping. First, the expression needs to be rewritten as $10n^{2}+an+bn-6$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-11$$ $$ab=10\left(-6\right)=-60$$
Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product $-60$.
Rewrite $10n^{2}-11n-6$ as $\left(10n^{2}-15n\right)+\left(4n-6\right)$.
$$\left(10n^{2}-15n\right)+\left(4n-6\right)$$
Factor out $5n$ in the first and $2$ in the second group.
$$5n\left(2n-3\right)+2\left(2n-3\right)$$
Factor out common term $2n-3$ by using distributive property.
$$\left(2n-3\right)\left(5n+2\right)$$
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 $ax^{2}+bx+c=0$.
$$10n^{2}-11n-6=0$$
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 $n=\frac{11±19}{20}$ when $±$ is plus. Add $11$ to $19$.
$$n=\frac{30}{20}$$
Reduce the fraction $\frac{30}{20}$ to lowest terms by extracting and canceling out $10$.
$$n=\frac{3}{2}$$
Now solve the equation $n=\frac{11±19}{20}$ when $±$ is minus. Subtract $19$ from $11$.
$$n=-\frac{8}{20}$$
Reduce the fraction $\frac{-8}{20}$ to lowest terms by extracting and canceling out $4$.
$$n=-\frac{2}{5}$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $\frac{3}{2}$ for $x_{1}$ and $-\frac{2}{5}$ for $x_{2}$.
Multiply $\frac{2n-3}{2}$ times $\frac{5n+2}{5}$ by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
