Question

Could someone please help me solve:

y(3y+7)=0

Thank you

Answer 1

if you start by distributing the y you get\[3y ^{2}+7y=0\]
then do you know what to do?

Answer 2

No, but if I remembered anything from these problems, distributing the y would have been the only thing I remembered :/

Answer 3

this is in the form \[ay ^{2}+by+c\] where c is a constant. in the case of\[3y ^{2}+7y=0\] your "c" value is zero. Then are you familiar with the quadratic formula?

Answer 4

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

Answer 5

does that help?

Answer 6

I think so,I'm not used to the program for the equations, so I'll see on paper :)

Answer 7

Yeah I have been on here for about two years and am just learning it.

Answer 8

haha :) I wrote it on paper, and it's starting to seem familiar :)

Answer 9

\[\frac{ \pm 7 \sqrt{7} }{ 3 }\] is this close?

Answer 10

all you have to do now is recognize that a=3, b=7, and c=0 because in the form \[ay ^{2}+by+c\] a is the 3, b is the 7 and anything plus the constant zero is just the same number so we don't write plus zero. It would be like saying I have five plus zero dollars. so the "c" in the case of\[3y ^{2}+7y=0 \] is zero because\[3y ^{2}+7y=0\]is the same as\[3y ^{2}+7y+0=0\]

Answer 11

\[a=3,b=7,c=0\]
\[\frac{ -b \pm \sqrt{b ^{2}-4ac} }{ 2a }=\frac{ -7\pm \sqrt{7^{2}-(4)(3)(0)} }{ (2)(3) }=\frac{ -7\pm \sqrt{49-0} }{ 6 }\]
\[=\frac{ -7-\sqrt{49} }{ 6 } and \frac{ -7+\sqrt{49} }{ 6 }=\frac{ 0 }{ 6 } and \frac{ -14 }{ 6 }\]

Answer 12

Yes! ok, I get it, thank you :)

Answer 13

Your welcome :)