Asked 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^2-4=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^2-4=0 or
6n^2-5n-4=0 does it factor to
(3n -4)(2n+ 1)
4y^2+12y+9=0.
(2y+3)^2
The first equation is -5n+6n^2-4, not 6n^2-5n-4=0. Thnx for all your help!!
-5n+6n^2-4=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^2-4=0 or
6n^2-5n-4=0 does it factor to
(3n -4)(2n+ 1)
4y^2+12y+9=0.
(2y+3)^2
The first equation is -5n+6n^2-4, not 6n^2-5n-4=0. Thnx for all your help!!
Answers
Answered by
mani
the roots of the equation are
n=[(-b)+sqrt(b^2-4*a*c)]/2*a
n=[(-b)-sqrt(b^2-4*a*c)]/2*a
thus,
n=[5+sqrt(25+96)]/2*6
n=[5-sqrt(25+96)]/2*6
so,the values of n for this eqn. are
n=4/3,-1/2
n=[(-b)+sqrt(b^2-4*a*c)]/2*a
n=[(-b)-sqrt(b^2-4*a*c)]/2*a
thus,
n=[5+sqrt(25+96)]/2*6
n=[5-sqrt(25+96)]/2*6
so,the values of n for this eqn. are
n=4/3,-1/2
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