How do you solve #x^2 + 6x – 7 = 0# using the quadratic formula?

Question

19934 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-14T08:28:01

#x=1#
#x=-7#

Given-

#x^2+7x-7=0#
#x^2+7x-x-7=0#
#x(x+7)-1(x+7)=0#
#(x-1)(x+7)#
#x-1=0#
#x=1#

#x+7=0#
#x=-7#

Answer 2 · 2016-04-14T08:30:14

The solutions are:

# x = 1 #

# x = -7#

#x^2 + 6x - 7 =0 #

The equation is of the form #color(blue)(ax^2+bx+c=0# where:

#a=1, b=6, c= - 7 #

The Discriminant is given by:

#Delta=b^2-4*a*c#

# = (6)^2-(4* 1 * (- 7))#

# = 36 + 28 = 64#

The solutions are normally found using the formula
#x=(-b+-sqrtDelta)/(2*a)#

#x = ((-6) +-sqrt(64))/(2*1) = (-6 +- 8)/2#

# x = (-6 + 8)/2 = 2 /2 = 1 #

# x = (-6 - 8)/2 = -14 /2 = -7#