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

Question

17792 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-08T13:59:50

#x=1,-3#

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#