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

Question

Eva Adamson · Accepted Answer

#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#

Abigail Allen · Answer

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#

Genesis Allen · Answer

undefined To solve the equation $ x^2 + 6x - 7 = 0 $ using the quadratic formula, follow these steps:

1. Identify the coefficients $ a = 1 $, $ b = 6 $, and $ c = -7 $.
2. Substitute these values into the quadratic formula: $ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} $.
3. Calculate the discriminant $ b^2 - 4ac = 6^2 - 4(1)(-7) = 36 + 28 = 64 $.
4. Take the square root of the discriminant: $ \sqrt{64} = 8 $.
5. Plug the values into the formula: $ x = \frac{{-6 \pm 8}}{2(1)} $.
6. Calculate the two possible solutions: $ x_1 = \frac{{-6 + 8}}{2} = \frac{2}{2} = 1 $ and $ x_2 = \frac{{-6 - 8}}{2} = \frac{-14}{2} = -7 $.
7. The solutions to the equation are $ x = 1 $ and $ x = -7 $.