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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
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

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nestor18Best ResponseYou've already chosen the best response.1
\[3y ^{2}2y5\] \[(2)\pm \sqrt{(2)^{2}4(3)(5)}/(2(3))\] your mistake is squaring the first coefficient. You do not need to. Recalculate from here, it should work out now.
 one year ago

nestor18Best ResponseYou've already chosen the best response.1
sorry also (2) is positive 2 as well.
 one year ago

Geometry_HaterBest ResponseYou've already chosen the best response.0
ok so my A coefficient is just 3
 one year ago

Geometry_HaterBest ResponseYou've already chosen the best response.0
Ok now what do i do in the equation?
 one year ago

nestor18Best ResponseYou've already chosen the best response.1
I also made the mistake of 3 in the equation it is actually 3
 one year ago

nestor18Best ResponseYou've already chosen the best response.1
I'll equate it out from the original
 one year ago

nestor18Best ResponseYou've already chosen the best response.1
\[(2\pm \sqrt{64})/6\] \[2\pm8/6\] =28=6/6=1 or 10/6=5/3
 one year ago
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