Question 1027738

Solve using quadratic formula.

1 over X^2 minus 3 equals 8 over x

The solution is -4 over 3 (plus or minus) the square root of 19 over 3. I don't understand how.
<pre>{{{1/x^2 - 3 = 8/x}}}
{{{1 - 3x^2 = 8x}}} -------- Multiplying by LCD, x<sup>2</sup>
{{{3x^2 + 8x - 1 = 0}}}
a = 3          b = 8           c = - 1
Quadratic equation formula: {{{x = (- b +- sqrt(b^2 - 4ac))/(2a)}}}
{{{x = (- 8 +- sqrt(8^2 - 4(3)(- 1)))/(2 * 3)}}} ------- Substituting 3 for a, 8 for b, and - 1 for c
{{{x = (- 8 +- sqrt(64 + 12))/6}}}
{{{x = (- 8 +- sqrt(76))/6}}}
{{{x = (- 8 +- sqrt(4 * 19))/6}}}
{{{x = (- 8 +- sqrt(4) * sqrt(19))/6}}}
{{{x = (- 8 +- 2sqrt(19))/6}}}
{{{x = 2(- 4 +- sqrt(19))/2(3)}}}
{{{x = cross(2)(- 4 +- sqrt(19))/cross(2)(3)}}}
{{{highlight(highlight_green(highlight(x = - 4/3 +- sqrt(19)/3)))}}}