Solve for \(x\)
\(4x^2=29x-7\)
\begin{align}4x^2-29x+7 & =0\\ 4x^2-28x-x+7 & = 0\\ 4x(x-7)-(x-7) & =0\\ (4x-1)(x-7) & = 0\\ x= \frac{1}{4} \quad or \quad x=7 \end{align}
\(7x=15-2x^2\)
\begin{align}2x^2+7x-15 & =0\\ 2x^2+10x-3x-15 & = 0\\ 2x(x+5)-3(x+5) & = 0\\ (2x-3)(x+5) & = 0\\ x=\frac{3}{2} \quad or \quad x=-5\\ \end{align}
\(6x-1=\frac{1}{x}\)
\begin{align}6x^2-x & =1\\ 6x^2-x-1 & =0\\ 6x^2 -3x+2x-1 & =0\\ 3x(2x-1)+(2x-1) & = 0\\ (3x+1)(2x-1) & =0\\ x= -\frac{1}{3} \quad or \quad x=\frac{1}{2}\\ \end{align}
\(x= \frac{2}{10x}-\frac{19}{10}\)
\begin{align}10x & = \frac{2}{x}-19\\ 10x^2 & = 2-19x\\ 10x^2+19x-2 & =0\\ 10x^2 +20x-x-2 & =0\\ 10x(x+2)-(x+2) & =0\\ (10x-1)(x+2) & =0\\ x=\frac{1}{10} \quad or \quad x=-2 \end{align}