Question

i need help figuring out how to solve -10y^3+25y^2+30y^4=0...been stuck on this problem for a half hour plz help

Answer 1

5y^2(2y+5+6y^2)=0 5y^2=0 implies y=0

so find the solutions to 6y^2+2y+5=0 and you are done

Answer 2

thats where im stuck at

Answer 3

i do not understand this at all a whole test of problems like this

Answer 4

use quadratic formula or you can use completing the square

Answer 5

hmm confused lol suck at math

Answer 6

\[y=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

Answer 7

our equation is in the form ax^2+bx+c=0
what is a?
what is b?
what is c?

Answer 8

and just pretend those x's are y's

Answer 9

o ic

Answer 10

ty trying to figure it all out

Answer 11

a=6
b=2
c=5
so we have
\[y=\frac{-2 \pm \sqrt{2^2-4(6)(5)}}{2(6)}=\frac{-2 \pm \sqrt{4-120}}{12}=\frac{-2 \pm \sqrt{-116}}{12}\]
\[=\frac{-2 \pm \sqrt{-1*4*29}}{12}=\frac{-2 \pm \sqrt{-1}\sqrt{4}\sqrt{29}}{12}=\frac{-2 \pm i*2*\sqrt{29}}{12}\]
\[=\frac{-2}{12} \pm i \frac{2\sqrt{29}}{12}=\frac{-1}{6} \pm i \frac{\sqrt{29}}{6}\]

Answer 12

tyvm

Answer 13

np