Question 655401
<pre>
We can get points on the parabola, say these 6, 
by substituting x=-4, x=-3, x=-2, x=-1, x=0, x=1

(-4,3), (-3,-1), (-2,-3), (-1,-3), (0,-1), (1,3),  

{{{drawing(400,400,-6.5,4.5,-5,6, graph(400,400,-6.5,4.5,-5,6), 
circle(-4,3,.1),circle(-3,-1,.1),circle(-2,-3,.1),circle(-1,-3,.1),circle(0,-1,.1),circle(1,3,.1) )}}} 

We sketch in the parabola:

{{{drawing(400,400,-6.5,4.5,-5,6, graph(400,400,-6.5,4.5,-5,6,x^2+3x-1),
 circle(-4,3,.1),circle(-3,-1,.1),circle(-2,-3,.1),circle(-1,-3,.1),circle(0,-1,.1),circle(1,3,.1) )}}}

We want to know where this parabola crosses the x-axis, which I
have marked below with red circles:

{{{drawing(400,400,-6.5,4.5,-5,6, graph(400,400,-6.5,4.5,-5,6,x^2+3x-1),
red(circle(-3.30277564,0,.3), circle(.30277564,0,.3)), 
circle(-4,3,.1),circle(-3,-1,.1),circle(-2,-3,.1),circle(-1,-3,.1),circle(0,-1,.1),circle(1,3,.1) )}}}

From the graph it looks like the x-intercept on the left crosses somewhere
between -3 and -4, and the x-intercept on the right crosses somewhere
between 0 and 1.

To get these accurately we use the quadratic formula to find the values
of x that cause y to be 0:

The solution to

ax² + bx + c = 0

is found by:

{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}

Your problem is 

x² + 3x - 1 = 0

So a=1, b=3, c=-1

{{{x = (-3 +- sqrt( 3^2-4*1*(-1) ))/(2*1) }}}

{{{x = (-3 +- sqrt(9+4))/2 }}}

{{{x = (-3 +- sqrt(13))/2 }}}

Use a calculator with the +

{{{x = (-3 + sqrt(13))/2 }}} = 0.3027756377, which is the x-coordinate
of the point circled on the right, and it is between 0 and 1.  So
that x-intercept on the right is

(0.3027756377, 0)

Use a calculator with the -

{{{x = (-3 - sqrt(13))/2 }}} = -3.302775638, which is the x-coordinate
of the point circled on the left, and it is between -3 and -4.  So
that x-intercept on the left is

(-3.302775638, 0)

Or you can write them in radical (exact) form:

({{{(-3 + sqrt(13))/2 }}}, 0) and ({{{(-3 - sqrt(13))/2 }}}, 0)

 Edwin</pre>