Quadratic Equations
posted by Emily .
Ok, I am supposed to find out what the variables are by factoring and finding square roots. I couldn't find the right factors of a and c that added up to b in the equation (did that make sense?). Please help me find them! TYVM
5n+6n^24=0
6y^2+12y+13=2y^2+4 simplified is:
4y^2+12y+9=0.
I know the answer to the second equation is 3/2, so I know it is a binomial square we are dealing with. Plz help, any help is greatly appreciated!
5n+6n^24=0 or
6n^25n4=0 does it factor to
(3n 4)(2n+ 1)
4y^2+12y+9=0.
(2y+3)^2
The first equation is 5n+6n^24, not 6n^25n4=0. Thnx for all your help!!

the roots of the equation are
n=[(b)+sqrt(b^24*a*c)]/2*a
n=[(b)sqrt(b^24*a*c)]/2*a
thus,
n=[5+sqrt(25+96)]/2*6
n=[5sqrt(25+96)]/2*6
so,the values of n for this eqn. are
n=4/3,1/2