How do I find the X coordinates of this calculus problem?

Question

I have some homework due tomorrow and don't have my book yet for the class because my school was out, so I can't use it for reference. Here is the problem.

Use zero (or root) to approximate the smaller of the two x-intercepts of each equation. Round to the nearest decimal.

1) y=x2+4x-3 (x2 is supposed to be x squared)

I was showed how to solve the problem on our graphing calculator but must have forgotten. Someone help me please!!

Answer 1

use the quadratic formula x = [ -b ± √ ( b² - 4 a c) ] / 2a.....{ -4 - √28} / 8...or complete the square y = x² + 4x + 4 - 7= (x+2)² -7----> x = -2 ± √ 7 and √7 ≈ 2.7

Answer 2

This is a quadratic, so you can always use the quadratic equation to find the roots (or zeros) of it.

[-4 +/- sqrt(16 + 12)  ] / 2

gives you two roots: [-4 + sqrt(28)] / 2 and [-4 - sqrt(28)] / 2

which ends up being .645 (so I guess round to .6) and -4.64 (so I guess round to -4.6).

And those are the answers.  

Hope that helps.