How do you solve #2x^2-1=7x# using the quadratic formula?

1 Answer
Mar 11, 2016

Answer:

#x=(7-sqrt57)/4# and#xx=(7+sqrt57)/4#

Explanation:

#2x^2-1=7x#
#=>2x^2-7x-1=0#
Dividing bothsides by 2
#=>x^2-7/2x-1/2=0#
#=>x^2-2*x*7/4+ (7/4)^2-(7/4)^2-1/2=0#
#=> (x-7/4)^2-49/16-1/2=0#
#=> (x-7/4)^2-(49+8)/16=0#
#=> (x-7/4)^2-(57)/16=0#
#=> (x-7/4+(sqrt57)/4)(x-7/4-(sqrt57)/4)=0#
#:.# #(x=7/4-sqrt57/4)=>x=(7-sqrt57)/4#

AND
#(x=7/4+sqrt57/4)=>x=(7+sqrt57)/4#