Factor out the common term \(x\).
\[\frac{x(2x+1)}{6}+\frac{2-3x}{4}=\frac{{x}^{2}-6}{12}\]
Multiply both sides by \(12\) (the LCM of \(6, 4, 12\)).
\[2x(2x+1)+3(2-3x)={x}^{2}-6\]
Expand.
\[4{x}^{2}+2x+6-9x={x}^{2}-6\]
Simplify \(4{x}^{2}+2x+6-9x\) to \(4{x}^{2}-7x+6\).
\[4{x}^{2}-7x+6={x}^{2}-6\]
Move all terms to one side.
\[4{x}^{2}-7x+6-{x}^{2}+6=0\]
Simplify \(4{x}^{2}-7x+6-{x}^{2}+6\) to \(3{x}^{2}-7x+12\).
\[3{x}^{2}-7x+12=0\]
Use the Quadratic Formula.
In general, given \(a{x}^{2}+bx+c=0\), there exists two solutions where:
\[x=\frac{-b+\sqrt{{b}^{2}-4ac}}{2a},\frac{-b-\sqrt{{b}^{2}-4ac}}{2a}\]
In this case, \(a=3\), \(b=-7\) and \(c=12\).
\[{x}^{}=\frac{7+\sqrt{{(-7)}^{2}-4\times 3\times 12}}{2\times 3},\frac{7-\sqrt{{(-7)}^{2}-4\times 3\times 12}}{2\times 3}\]
Simplify.
\[x=\frac{7+\sqrt{95}\imath }{6},\frac{7-\sqrt{95}\imath }{6}\]
\[x=\frac{7+\sqrt{95}\imath }{6},\frac{7-\sqrt{95}\imath }{6}\]
x=(7+sqrt(95)*IM)/6,(7-sqrt(95)*IM)/6
