how do i solve this equation: 3x^2-1 is greater than or equal to (1/2) (3+x^2)the (1/2) in parentheses is a fraction

Solve `3x^2-1
>= (3+x^2)/2` .

First we rewrite as an
equivalent inequality (one that has the same solutions as the
original)

`3x^2-1 >=
(3+x^2)/2`

class="AM">`6x^2-2>=3+x^2` Multiply both sides by
2

`5x^2-5>=0`
Pull terms to left side by subtraction

class="AM">`5(x^2-1)>=0 => 5(x+1)(x-1)>=0`
Factor.

Now the left hand side is zero when x=1 or -1. To
find the intervals where it is greater than zero we plug in test values. The intervals
we are interested in are x<-1, -1<x<1, and x>1; so we try
x=-2,x=0, and x=2.

If x=-2 the left hand side is positive;
so if x<-1 then the left-hand side is greater than
zero.

If x=0 the left-hand side is negative so the
left-hand side is negative for -1<x<1.

If x=2
the LHS is positive so LHS is greater than zero if
x>1.

We can also look at the graph to check our work
-- notice that the "skinnier" parabola is above the wider parabola as long as
x<-1 or x>1.

src="/jax/includes/tinymce/jscripts/tiny_mce/plugins/asciisvg/js/d.svg"
sscr="-7.5,7.5,-5,5,1,1,1,1,1,300,200,func,3x^2-1,null,0,0,,,black,1,none,func,(3+x^2)/2,null,0,0,,,black,1,none"/>

Thus
the solution to `3x^2-1 >= (3+x^2)/2` is
`x<= -1` or class="AM">`x>=1` .