Answer
$-\displaystyle \frac{1}{4}$ and $\displaystyle \frac{2}{3}$
Work Step by Step
Set the $RHS$ to 0 by subtracting 2 from both sides
$12x^{2}-5x=2\qquad.../-$2
$12x^{2}-5x-2=0$
To factor the trinomial, we search for
factors of $12\times(-2)=-24, $ whose sum is $-5:\qquad $
We find $-8$ and $+3$.
$12x^{2}-5x-2=12x^{2}-8x+3x-2$
... and factor in pairs:
$=4x(3x-2)+1(3x-2)$
$=(4x+1)(3x-2)$
Our equation becomes
$(4x+1)(3x-2)=0$
By the zero product principle,
$4x+1=0$ or $3x-2=0$
$4x=-1$ or $3x=2$
$x=-\displaystyle \frac{1}{4}$ or $x=\displaystyle \frac{2}{3}$
The solutions are $-\displaystyle \frac{1}{4}$ and $\displaystyle \frac{2}{3}$.
