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Geometry_Hater
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.
\[3y ^{2}-2y-5\] \[-(-2)\pm \sqrt{(-2)^{2}-4(-3)(-5)}/(2(3))\] your mistake is squaring the first coefficient. You do not need to. Recalculate from here, it should work out now.
sorry also -(-2) is positive 2 as well.
ok so my A coefficient is just 3
Ok now what do i do in the equation?
I also made the mistake of -3 in the equation it is actually 3
I'll equate it out from the original
\[(2\pm \sqrt{64})/6\] \[2\pm8/6\] =2-8=-6/6=-1 or 10/6=5/3