How do you solve #5n^2-10n+59=0#?

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Answer 1 · 2017-02-01T22:16:23

#x=1+-3/5sqrt30i#

If the general form of the given equation is:

#ax^2+2nx+c#,

then

#x=(-n+-sqrt((n)^2-ac))/a#

Therefore:

#x=(5+-sqrt(5^2-5*59))/5#

#=(5+-sqrt(25-295))/5#

The solutions are complex, since the number under the square root is negative, then

#x=(5+-sqrt(-270))/5=1+-3/5sqrt30i#