Question 825443: how would you put 4x^2+5x=(3x-1)^2-3x^2 in the standard ax^2+bx+c=0 form
Answer by TimothyLamb(4379) (Show Source):
You can put this solution on YOUR website! ---
4x^2 + 5x = (3x - 1)^2 - 3x^2
7x^2 + 5x = (3x - 1)^2
7x^2 + 5x = (3x - 1)(3x - 1)
7x^2 + 5x = 9x^2 - 6x + 1
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answer:
2x^2 - 11x + 1 = 0
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