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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.
 2 years ago
 2 years ago
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.
 2 years ago
 2 years ago

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nestor18 Group TitleBest 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.
 2 years ago

nestor18 Group TitleBest ResponseYou've already chosen the best response.1
sorry also (2) is positive 2 as well.
 2 years ago

Geometry_Hater Group TitleBest ResponseYou've already chosen the best response.0
ok so my A coefficient is just 3
 2 years ago

Geometry_Hater Group TitleBest ResponseYou've already chosen the best response.0
Ok now what do i do in the equation?
 2 years ago

nestor18 Group TitleBest ResponseYou've already chosen the best response.1
I also made the mistake of 3 in the equation it is actually 3
 2 years ago

nestor18 Group TitleBest ResponseYou've already chosen the best response.1
I'll equate it out from the original
 2 years ago

nestor18 Group TitleBest 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
 2 years ago

Geometry_Hater Group TitleBest ResponseYou've already chosen the best response.0
Thanks
 2 years ago
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