Answer
The solution set is $\left\{-\frac{1}{2}, \frac{1}{2}\right\}$.
Work Step by Step
Add $12x^2-3$ to both sides to find:
$0=12x^2-3$
Divide by $3$ on both sides to find:
$0=4x^2-1
\\0=(2x)^2-1^{2}$
RECALL:
A difference of two squares can be factored using the formula:
$a^2-b^2=(a-b)(a+b)$
Factor the difference of two squares to find:
$(2x-1)(2x+1)=0$
Equate each factor to 0 then solve each equation to find:
$2x-1=0 \text{ or } 2x+1 = 0\\
2x=1 \text{ or } 2x=-1
\\x=\frac{1}{2} \text{ or }x=-\frac{1}{2}$
The solution set is $\left\{-\frac{1}{2}, \frac{1}{2}\right\}$.
