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 $ax^{2}+bx+c=0$.
$$12x^{2}-15x+2=0$$
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 $x=\frac{15±\sqrt{129}}{24}$ when $±$ is plus. Add $15$ to $\sqrt{129}$.
$$x=\frac{\sqrt{129}+15}{24}$$
Divide $15+\sqrt{129}$ by $24$.
$$x=\frac{\sqrt{129}}{24}+\frac{5}{8}$$
Now solve the equation $x=\frac{15±\sqrt{129}}{24}$ when $±$ is minus. Subtract $\sqrt{129}$ from $15$.
$$x=\frac{15-\sqrt{129}}{24}$$
Divide $15-\sqrt{129}$ by $24$.
$$x=-\frac{\sqrt{129}}{24}+\frac{5}{8}$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $\frac{5}{8}+\frac{\sqrt{129}}{24}$ for $x_{1}$ and $\frac{5}{8}-\frac{\sqrt{129}}{24}$ for $x_{2}$.
