Another Algebra Problem. Help!

2050 = 3n [n+(10/3)]

how to get value of n......

distribute the multiplication on the right half and rearrange and you'll get a quadratic

2050 = 3n^2 + 10n

3n^2 +10n - 2050 = 0

Solve with the quadratic formula

2050 = 3n [n+(10/3)]

expand the brackets
2050 = 3n^2 + 10n

rewrite it as
3n^2 + 10n - 2050 = 0

use the quadratic formula
n = (-10 +/- sqrt(100 - 4*3*-2050))/6 = (-10 +/- sqrt(24700))/6 = 24.53 or -27.86

Your equation is equivalent to 3n^2+10n-2050=0. We can easily solve this quadratic equation.

2050 = 3n + (n + 3.3333)
2050 - n + 3 3333 = 3n
683.333 - n/3 = n
BOO! cba