## Geometry_Hater Group Title 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. one year ago one year ago

1. nestor18 Group Title

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

2. nestor18 Group Title

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

3. Geometry_Hater Group Title

ok so my A coefficient is just 3

4. nestor18 Group Title

Correct

5. Geometry_Hater Group Title

Ok now what do i do in the equation?

6. nestor18 Group Title

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

7. nestor18 Group Title

I'll equate it out from the original

8. nestor18 Group Title

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

9. nestor18 Group Title

y=

10. Geometry_Hater Group Title

Thanks