Solve for X using quadratic function

3y^2 - 2y = 5

I already subtracted the 5 to get 0 so my
A = 3^2
b= -2
c= -5
But im having trouble with the actual equation part which happens to be -2 + or - square root -2^2 - 4 * 9 * -5 over 2 * 9.

Question

Solve for X using quadratic function

3y^2 - 2y = 5

I already subtracted the 5 to get 0 so my 
A = 3^2 
b= -2
c= -5
But im having trouble with the actual equation part which happens to be -2 + or - square root -2^2 - 4 * 9 * -5 over 2 * 9.

anonymous · Answer

$$3y ^{2}-2y-5$$
$$-(-2)\pm \sqrt{(-2)^{2}-4(-3)(-5)}/(2(3))$$
your mistake is squaring the first coefficient. You do not need to. Recalculate from here, it should work out now.

anonymous · Answer

sorry also -(-2) is positive 2 as well.

geometry_hater · Answer

ok so my A coefficient is just 3

anonymous · Answer

Correct

geometry_hater · Answer

Ok now what do i do in the equation?

anonymous · Answer

I also made the mistake of -3 in the equation it is actually 3

anonymous · Answer

I'll equate it out from the original

anonymous · Answer

$$(2\pm \sqrt{64})/6$$
$$2\pm8/6$$
=2-8=-6/6=-1
or
10/6=5/3

anonymous · Answer

y=

geometry_hater · Answer

Thanks