Question 575503
Sometimes it simpifies things a bit by temporarily using a substitute variable:
Temporarily let {{{x^2 = y}}} so now you have:
{{{3y^2-11y-20 = 0}}} Factor:
{{{(3y+4)(y-5) = 0}}} Solve for y:
{{{3y+4 = 0}}} or {{{y-5 = 0}}} so...
{{{y = -4/3}}} or {{{y = 5}}} Replace {{{y = x^2}}} to get...
{{{x^2 = -4/3}}} or {{{x^2 = 5}}} Taking the square roots, we get...
{{{x = sqrt(-4/3)}}} or {{{x = -sqrt(-4/3)}}} or {{{x = sqrt(5)}}} or {{{x = -sqrt(5)}}}
If you need your answers in approximate form, then...
{{{x = 2.236}}} or {{{x = -2.236}}} or {{{x = 1.155i}}} or {{{x = -1.155i}}}
Here's what the graph looks like:
{{{graph(400,400,-5,5,-30,5,3x^4-11x^2-20)}}}