solve by completing the square 4x^2-x=0 and solve by factoring 3x^2+1=4x

solve by factoring 3x^2+1=4x

$$\small{\text{
\begin{array}{rcl}
$
3x^2+1 & = & 4x \\
3x^2 -4x + 1 & = &0 \\
(x-1)(3x-1) &=& 0 \\
3(x-1)(x-\frac{1}{3}) &=& 0
$
\end{array}
}}
\\
\small{\text{
$ x_1 = 1 \qquad x_2 = \frac{1}{3}
$
}}
\\
\small{\text{
Check: $\quad 3*1^2+1=4*1 $ okay $ \qquad 3*\frac{1}{3}*\frac{1}{3}+1=4*\frac{1}{3} $ okay
}}$$

4x^2-x=0

Please note: If the question was not so specific you would never so this one this way!

$$\\4x^2-x=0\\\\
x^2-\frac{1}{4}x=0\\\\
x^2-\frac{1}{4}x+(\frac{1}{2}\times\frac{1}{4})^2=0+(\frac{1}{2}\times\frac{1}{4})^2\\\\
x^2-\frac{1}{4}x+(\frac{1}{8})^2=0+(\frac{1}{8})^2\\\\
x^2-\frac{1}{4}x+\frac{1}{64}=\frac{1}{64}\\\\
x^2-\frac{1}{4}x+\frac{1}{64}=\frac{1}{64}\\\\
(x-\frac{1}{8})^2=\frac{1}{64}\\\\
x-\frac{1}{8}=\pm\frac{1}{8}\\\\
x=\frac{1}{8}+\frac{1}{8}\qquad or \qquad x=-\frac{1}{8}+\frac{1}{8}\\\\
x=\frac{1}{4}\qquad or \qquad x=0 \\\\$$

3x^2+1=4x

$$\\3x^2-4x+1=0\\\\
3x^2-3x-1x+1=0\\\\
3x(x-1)-1(x-1)=0\\\\
(3x-1)(x-1)=0\\\\
x=1/3 \qquad or \qquad x=1$$