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x^2+x/7-2/7 = 0

Add 2/7 to both sides:

x^2+x/7 = 2/7

Add 1/196 to both sides:

x^2+x/7+1/196 = 57/196

Write the left hand side as a square:

(x+1/14)^2 = 57/196

Take the square root of both sides:

x+1/14 = sqrt(57)/14 or x+1/14 = -sqrt(57)/14

Subtract 1/14 from both sides:

x = sqrt(57)/14-1/14 or x+1/14 = -sqrt(57)/14

Subtract 1/14 from both sides:

Answer: |x = sqrt(57)/14-1/14                 or            x = -1/14-sqrt(57)/14

http://www.cymath.com/answer.php?q=(3x%2B2)%2Fx%3Dx%2F(-2x%2B1)

(3x+2)/x=x/(-2x+1)

$\begin{array}{rcll} \dfrac{3x+2} {x} &=& \dfrac{x} {-2x+1} \\\\ \dfrac{3x+2} {x} &=& \dfrac{x} {1-2x} \\\\ 3 + \dfrac{2} {x} &=& \dfrac{1} { \frac{1-2x}{x} } \\\\ 3 + \dfrac{2} {x} &=& \dfrac{1} { \frac{1}{x} - 2 } \qquad & | \cdot \dfrac{1}{x} - 2\\\\ \left(\dfrac{1}{x} - 2 \right)\cdot \left(3 + \dfrac{2} {x} \right) &=& 1 \\\\ \left(\dfrac{1}{x} - 2 \right)\cdot \left(3 + \dfrac{2} {x} \right) &=& 1 \\\\ \left(\dfrac{1}{x} - 2 \right)\cdot \left(3 + 2\cdot \dfrac{1} {x} \right) &=& 1 \\\\ \end{array}$

We substitute:  $u=\dfrac{1}{x}$

$\begin{array}{rcll} \left(u - 2 \right)\cdot \left(3 + 2\cdot u \right) &=& 1 \\ 3u+2u^2-6-4u &=& 1 \\ 2u^2 -u-7 &=& 0 \\ \end{array}$

$\boxed{~ \begin{array}{rcll} ax^2+bx+c &=& 0 \\ x &=& \dfrac{-b \pm \sqrt{b^2-4ac} }{2a} \end{array} ~}$

$\begin{array}{rcll} 2u^2 -u-7 &=& 0 \qquad & a = 2 \qquad b = -1 \qquad c = -7\\ u_{1,2} &=& \dfrac{-(-1) \pm \sqrt{(-1)^2-4\cdot 2 \cdot (-7)} }{2\cdot 2}\\ u_{1,2} &=& \dfrac{1 \pm \sqrt{1+56} }{4}\\ u_{1,2} &=& \dfrac{1 \pm \sqrt{57} }{4}\\\\ u_1 &=& \dfrac{1 + \sqrt{57} }{4}\\ x_1 &=& \frac{1}{u_1} \\ \mathbf{x_1} &\mathbf{=}& \mathbf{\dfrac{4}{1 + \sqrt{57} } }\\\\ u_2 &=& \dfrac{1 - \sqrt{57} }{4}\\ x_2 &=& \frac{1}{u_2} \\ \mathbf{x_2} &\mathbf{=}& \mathbf{\dfrac{4}{1 - \sqrt{57} } } \end{array}$