SOLUTION: 'solve each equation' please. (: 3n^2+6=11n

Algebra ->  Equations -> SOLUTION: 'solve each equation' please. (: 3n^2+6=11n      Log On


   

Question 194069: 'solve each equation' please. (:
3n^2+6=11n
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
3n%5E2%2B6=11n Start with the given equation.


3n%5E2%2B6-11n=0 Subtract 11n from both sides.


3n%5E2-11n%2B6=0 Rearrange the terms.


Notice we have a quadratic equation in the form of an%5E2%2Bbn%2Bc where a=3, b=-11, and c=6


Let's use the quadratic formula to solve for n


n+=+%28-b+%2B-+sqrt%28+b%5E2-4ac+%29%29%2F%282a%29 Start with the quadratic formula


n+=+%28-%28-11%29+%2B-+sqrt%28+%28-11%29%5E2-4%283%29%286%29+%29%29%2F%282%283%29%29 Plug in a=3, b=-11, and c=6


n+=+%2811+%2B-+sqrt%28+%28-11%29%5E2-4%283%29%286%29+%29%29%2F%282%283%29%29 Negate -11 to get 11.


n+=+%2811+%2B-+sqrt%28+121-4%283%29%286%29+%29%29%2F%282%283%29%29 Square -11 to get 121.


n+=+%2811+%2B-+sqrt%28+121-72+%29%29%2F%282%283%29%29 Multiply 4%283%29%286%29 to get 72


n+=+%2811+%2B-+sqrt%28+49+%29%29%2F%282%283%29%29 Subtract 72 from 121 to get 49


n+=+%2811+%2B-+sqrt%28+49+%29%29%2F%286%29 Multiply 2 and 3 to get 6.


n+=+%2811+%2B-+7%29%2F%286%29 Take the square root of 49 to get 7.


n+=+%2811+%2B+7%29%2F%286%29 or n+=+%2811+-+7%29%2F%286%29 Break up the expression.


n+=+%2818%29%2F%286%29 or n+=++%284%29%2F%286%29 Combine like terms.


n+=+3 or n+=+2%2F3 Simplify.


So the answers are n+=+3 or n+=+2%2F3