How do you solve #2n ^ { 2} - 5n = 9#?

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Answer 1 · 2018-03-12T04:50:18

#n=(5+sqrt97)/4,# #(5-sqrt97)/4#

Solve:

#2n^2-5n=9#

Subtract #9# from both sides of the equation.

#2n^2-5n-9=0# is a quadratic equation in standard form:

#ax^2+bx+c=0,#

where:

#a=2,# #b=-5,# #c=-9#

Use the quadratic formula to solve the equation.

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

Substitute #n# for #x#. Plug in the known values and solve.

#n=(-(-5)+-sqrt((-5)^2-4*2*-9))/(2*2)#

Simplify.

#n=(5+-sqrt97)/4#

#97# is a prime number, so it cannot be factored further.

#n=(5+sqrt97)/4,# #(5-sqrt97)/4#