Answer
the solution set is\[\left\{ -\sqrt{5},\sqrt{5} \right\}\].
Work Step by Step
The given equation can be written as\[3{{x}^{2}}-6x-3=12-6x\].
Simplify this as follows:
\[\begin{align}
& 3{{x}^{2}}-6x-3=12-6x \\
& 3{{x}^{2}}-6x-3-12+6x=0 \\
& 3{{x}^{2}}-15=0
\end{align}\]
Compare the given equation with the equation\[a{{x}^{2}}+bx+c=0\], where\[a=3,\ b=0,\ \text{and }c=-15\].
Now, put these values in the quadratic formula \[x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
\[\begin{align}
& x=\frac{-0\pm \sqrt{{{0}^{2}}-4\times \left( 3 \right)\times \left( -15 \right)}}{2\times 3} \\
& =\frac{0\pm \sqrt{0+180}}{6} \\
& =\frac{0\pm 6\sqrt{5}}{6} \\
& =-\sqrt{5},\sqrt{5}
\end{align}\]
Hence, the solution set is\[\left\{ -\sqrt{5},\sqrt{5} \right\}\].
