How do you solve #2n ^ { 2} - 5= 5n#?

Question

617 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-28T02:12:07

#n=(5+sqrt65)/4,##(5-sqrt65)/4#

Solve:

#2n^2-5=5n#

Move all terms to the left side and set them equal to zero.

#2n^2-5n-5=0# is a quadratic equation in standard form:

#ax^2+bx+c#,

where #a=2#, #b=-5#, #c=-5#

You can use the quadratic formula to find the solutions for #n#.

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

Substitute #n# for #x# and plug the known values.

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

Simplify.

#n=(5+-sqrt(65))/4#

The prime factors of #65# are #5# and #13#, therefore it cannot be simplified.

Solutions for #n#.

#n=(5+sqrt65)/4,##(5-sqrt65)/4#