Answer
no real solution
Work Step by Step
In the form $ax^2+bx+c,$ the given equation, $
3x^2+1=x
,$ is equivalent to
\begin{align*}
3x^2-x+1&=0
.\end{align*}
In the equation above, $a=
3
,$ $b=
-1
,$ and $c=
1
.$ Using the Quadratic Formula which is given by $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a},$ then
\begin{align*}
x&=\dfrac{-(-1)\pm\sqrt{(-1)^2-4(3)(1)}}{2(3)}
\\\\&=
\dfrac{1\pm\sqrt{1-12}}{6}
\\\\&=
\dfrac{1\pm\sqrt{-11}}{6}
.\end{align*}
Since the radicand is negative (i.e. $-11$), there is no real solution.
