How do you solve #x^2-7x-5=0# using quadratic formula?

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Answer 1 · 2016-11-15T03:38:18

#x=7.65 and x=-0.65#

#x^2-7x-5=0#

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

Values given #a=1, b=-7, c=-5#

#x=[-(-7)+-sqrt((-7)^2-4xx1xx(-5))]/(2xx1)#

#x=[7+-sqrt(49+20)]/2#

#x=[7+-sqrt(69)]/2#

#x=[7+sqrt(69)]/2 and x=[7-sqrt(69)]/2#

#:.x=7.65 and x=-0.65#