How do you factor #n^2+5n+7#?

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Answer 1 · 2015-10-01T21:56:41

#n=((-5+sqrt(3)i)/2),((-5-sqrt(3)i)/2)#

#n^2+5n+7# is a quadratic equation #ax^2+bx+c#, where #a=1, b=5, c=7#.

You can use the quadratic formula to factor this quadratic equation.

Quadratic Formula

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

Substitute #n# for #x#.

#n=(-5+-sqrt(5^2-4*1*7))/(2*1)=#

Simplify.

#n=(-5+-sqrt(25-28))/2=#

Simplify.

#n=(-5+-sqrt(-3))/2=#

Simplify.

#n=(-5+-sqrt(3)i)/2=#

#n=((-5+sqrt(3)i)/2),((-5-sqrt(3)i)/2)#