How do you solve #n^ { 2} + 18n - 34= 0#?

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Answer 1 · 2017-03-10T20:10:41

The solutions are #S={-19.72, 1.72}#

We compare this equation to

#ax^2+bx+c=0#

#n^2+18n-34=0#

We calculate the discriminant

#Delta=b^2-4ac=18^2-4*1*(-34)#

#=324+136=460#

As #Delta>0#, there are 2 real solutions

#n=(-b+-sqrt Delta)/(2a)#

#=(-18+-sqrt460)/2#

#n_1=(-18+21.448)/2=1.72#

#n_2=(-18-21.448)/2=-19.72#