How do you solve #14n^2 + 24n - 5004 = 0#?

2 Answers
Mar 5, 2018

Explanation:

The straightforward way is to use the quadratic formula:
#n=(-b+-sqrt(b^2-4ac))/(2a)#
This is for expressions of the form #an^2 + bn + c#, so in your case a=14, b=24, c=-5004 .

A slightly more complicated way is to reverse foil, trying to factor out your a, b, and c so you can write you equation into
#(xn + y)(sn +t) = 0#, where x, y, s, and t are all numbers. Then you can solve one #()# at a time. But if you have a calculator, the quadratic formula is the best way to go. It'll work every time.

Mar 5, 2018

Answer:

#n = -6/7+-15/7sqrt(78)#

Explanation:

Complete the square then use the difference of squares identity to find:

#0 = 7/2(14n^2+24n-5004)#

#color(white)(0) = 49n^2+84n-17514#

#color(white)(0) = (7n)^2+2(7n)(6)+36-17550#

#color(white)(0) = (7n)^2+2(7n)(6)+6^2-(15^2 * 78)#

#color(white)(0) = (7n+6)^2-(15sqrt(78))^2#

#color(white)(0) = ((7n+6)-15sqrt(78))((7n+6)+15sqrt(78))#

#color(white)(0) = (7n+6-15sqrt(78))(7n+6+15sqrt(78))#

Hence:

#7n = -6+-15sqrt(78)#

So:

#n = -6/7+-15/7sqrt(78)#