How do you solve x^2+2x-3=0 using the quadratic formula?

1 Answer
Mar 8, 2018

x=1,-3

Explanation:

We have x^2+2x-3=0. For any ax^2+bx+c=0, x is given by:

x=(-b+-sqrt(b^2-4ac))/(2a)

Here, a=1, b=2, c=-3. Inputting:

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

x=(-2+-sqrt(4+12))/2

x=(-2+-sqrt(16))/2

x=(-2+-4)/2

x=2/2,-6/2

x=1,-3