Combine $6a^{2}$ and $-81a^{2}$ to get $-75a^{2}$.
$$factor(-75a^{2}-7a+10a+5)$$
Combine $-7a$ and $10a$ to get $3a$.
$$factor(-75a^{2}+3a+5)$$
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 $ax^{2}+bx+c=0$.
$$-75a^{2}+3a+5=0$$
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 $a=\frac{-3±\sqrt{1509}}{-150}$ when $±$ is plus. Add $-3$ to $\sqrt{1509}$.
$$a=\frac{\sqrt{1509}-3}{-150}$$
Divide $-3+\sqrt{1509}$ by $-150$.
$$a=-\frac{\sqrt{1509}}{150}+\frac{1}{50}$$
Now solve the equation $a=\frac{-3±\sqrt{1509}}{-150}$ when $±$ is minus. Subtract $\sqrt{1509}$ from $-3$.
$$a=\frac{-\sqrt{1509}-3}{-150}$$
Divide $-3-\sqrt{1509}$ by $-150$.
$$a=\frac{\sqrt{1509}}{150}+\frac{1}{50}$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $\frac{1}{50}-\frac{\sqrt{1509}}{150}$ for $x_{1}$ and $\frac{1}{50}+\frac{\sqrt{1509}}{150}$ for $x_{2}$.
