How do you solve $\frac{3}{2 x} - \frac{9}{2} = 6 x$ $3/(2x)-9/2=6x$ ?

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Answer 1 · 2018-05-30T19:40:29

$x = \frac{1}{4}, - 1$ $x=1/4, -1$

$\frac{3}{2 x} - \frac{9}{2} = 6 x$ $3/(2x)-9/2=6x$

$\frac{3}{2 x} - \frac{9 x}{2 x} = 6 x$ $3/(2x)-(9x)/(2x)=6x$

$\frac{3 - 9 x}{2 x} = 6 x$ $(3-9x)/(2x)=6x$

$3 - 9 x = 12 x^{2}$ $3-9x=12x^2$

$12 x^{2} + 9 x - 3 = 0$ $12x^2+9x-3 = 0$

$4 x^{2} + 3 x - 1 = 0$ $4x^2+3x-1=0$

$(4 x - 1) (x + 1) = 0$ $(4x-1)(x+1)=0$

$x = \frac{1}{4}, - 1$ $x=1/4, -1$

How do you solve 32x−92=6x3/(2x)-9/2=6x?