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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.

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  1. nestor18
    • 3 years ago
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    \[3y ^{2}-2y-5\] \[-(-2)\pm \sqrt{(-2)^{2}-4(-3)(-5)}/(2(3))\] your mistake is squaring the first coefficient. You do not need to. Recalculate from here, it should work out now.

  2. nestor18
    • 3 years ago
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    sorry also -(-2) is positive 2 as well.

  3. Geometry_Hater
    • 3 years ago
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    ok so my A coefficient is just 3

  4. nestor18
    • 3 years ago
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    Correct

  5. Geometry_Hater
    • 3 years ago
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    Ok now what do i do in the equation?

  6. nestor18
    • 3 years ago
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    I also made the mistake of -3 in the equation it is actually 3

  7. nestor18
    • 3 years ago
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    I'll equate it out from the original

  8. nestor18
    • 3 years ago
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    \[(2\pm \sqrt{64})/6\] \[2\pm8/6\] =2-8=-6/6=-1 or 10/6=5/3

  9. nestor18
    • 3 years ago
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    y=

  10. Geometry_Hater
    • 3 years ago
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    Thanks

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