How do you solve #x^2 + 2x-7=0#?

1 Answer
Apr 19, 2016

#x = -1 pm 2sqrt{2}#

Explanation:

Use the quadratic formula for this question where you can't find a master product that will have factors that add to the #b# term.

Quadratic Formula
#x = (-(b) pm sqrt{b^2 - 4ac})/(2a)#

Substitute the values into its corresponding places and evaluate.
#x = (-(2) pm sqrt{2^2 - 4(1)(-7)})/(2(1))#
#x = (-2 pm sqrt{4 + 28})/(2)#
#x = (-2 pm sqrt{32})/(2)#
#x = (-2 pm 4sqrt{2})/(2)#

Simplify the terms (other than the 2 inside of the radical sign) by diving them by the common factor of 2.
#x = (-1 pm 2sqrt{2})/(1)#

Answer: #x = -1 pm 2sqrt{2}#