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Geometry_Hater
 3 years 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.
Geometry_Hater
 3 years 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.

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[3y ^{2}2y5\] \[(2)\pm \sqrt{(2)^{2}4(3)(5)}/(2(3))\] your mistake is squaring the first coefficient. You do not need to. Recalculate from here, it should work out now.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sorry also (2) is positive 2 as well.

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I also made the mistake of 3 in the equation it is actually 3

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I'll equate it out from the original

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[(2\pm \sqrt{64})/6\] \[2\pm8/6\] =28=6/6=1 or 10/6=5/3
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