Solve for \(x\) in \({x}^{2}-4x+12=0\).
Solve for \(x\).
\[{x}^{2}-4x+12=0\]
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, \(a=1\), \(b=-4\) and \(c=12\).
\[{x}^{}=\frac{4+\sqrt{{(-4)}^{2}-4\times 12}}{2},\frac{4-\sqrt{{(-4)}^{2}-4\times 12}}{2}\]
Simplify.
\[x=\frac{4+4\sqrt{2}\imath }{2},\frac{4-4\sqrt{2}\imath }{2}\]
\[x=\frac{4+4\sqrt{2}\imath }{2},\frac{4-4\sqrt{2}\imath }{2}\]
Simplify solutions.
\[x=\frac{4(1+\sqrt{2}\imath )}{2},\frac{4(1-\sqrt{2}\imath )}{2}\]
\[x=\frac{4(1+\sqrt{2}\imath )}{2},\frac{4(1-\sqrt{2}\imath )}{2}\]
Substitute \(x=\frac{4(1+\sqrt{2}\imath )}{2},\frac{4(1-\sqrt{2}\imath )}{2}\) into \(3{x}^{2}+12x+12=0\).
Start with the original equation.
\[3{x}^{2}+12x+12=0\]
Let \(x=\frac{4(1+\sqrt{2}\imath )}{2},\frac{4(1-\sqrt{2}\imath )}{2}\).
\[3{(\frac{4(1+\sqrt{2}\imath )}{2},\frac{4(1-\sqrt{2}\imath )}{2})}^{2}+12(\frac{4(1+\sqrt{2}\imath )}{2},\frac{4(1-\sqrt{2}\imath )}{2})+12=0\]
Simplify.
\[\frac{48{(1+\sqrt{2}\imath )}^{2}}{{(2,8(1-\sqrt{2}\imath ))}^{2}}+\frac{48(1+\sqrt{2}\imath )}{2,8(1-\sqrt{2}\imath )}+12=0\]
\[\frac{48{(1+\sqrt{2}\imath )}^{2}}{{(2,8(1-\sqrt{2}\imath ))}^{2}}+\frac{48(1+\sqrt{2}\imath )}{2,8(1-\sqrt{2}\imath )}+12=0\]
Since \(\frac{48{(1+\sqrt{2}\imath )}^{2}}{{(2,8(1-\sqrt{2}\imath ))}^{2}}+\frac{48(1+\sqrt{2}\imath )}{2,8(1-\sqrt{2}\imath )}+12=0\) is not true, this is an inconsistent system.
