How do you solve #2x ^ { 2} - 5x + 1= 0#?

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Answer 1 · 2016-11-29T04:23:25

#(5 + sqrt(17)) / 4 = 2.2807764064044#
#(5 - sqrt(17)) / 4 = 0.21922359359558#

I always try to find a reverse FOIL answer, but in this case, the quadratic equation is required.

#(-b +- sqrt(b^2 - 4ac)) / (2a)#

In this equation:
#a = 2#
#b = -5#
#c = 1#

#-(-5) +-sqrt((-5)^2 - 4(2)(1)) / (2(2))#

#(5 + sqrt(25 - 8)) / 4#
#(5 + sqrt(17)) / 4 = 2.2807764064044#

#(5 - sqrt(25 - 8)) / 4#
#(5 - sqrt(17)) / 4 =0.21922359359558#