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4y^(2)+2y-(k-4)=0

4y2+2y−(k−4)=0 4 y^{2}+2 y-(k-4)=0

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Full solution

Q. 4y2+2y−(k−4)=0 4 y^{2}+2 y-(k-4)=0
  1. Move constant term: Step Title: Move the constant term to the other side\newlineCalculation: Add k−4k - 4 to both sides to get 4y2+2y+k−4=04y^2 + 2y + k - 4 = 0.

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