How do you write $21 x^{2} + 5 x - 35 = 3 x^{2} - 4 x$ $21x^2+5x-35=3x^2-4x$ in standard form?

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Answer 1 · 2016-12-07T04:00:35

$18 x^{2} + 9 x - 35 = 0$ $18x^2 + 9x - 35 = 0$

To get to standard form you must all of the terms must be on one side of the equation and the quadratic must be equal to $0$ $0$ :

$21 x^{2} + 5 x - 35 - 3 x^{2} + 4 x = 3 x^{2} - 4 x - 3 x^{2} + 4 x$ $21x^2 + 5x - 35 - 3x^2 + 4x = 3x^2 - 4x - 3x^2 + 4x$

$21 x^{2} - 3 x^{2} + 5 x + 4 x - 35 = 3 x^{2} - 3 x^{2} - 4 x + 4 x$ $21x^2 - 3x^2 + 5x + 4x - 35 = 3x^2 - 3x^2 - 4x + 4x$

$(21 - 3) x^{2} + (5 + 4) x - 35 = 0 - 0$ $(21 - 3)x^2 + (5 + 4)x - 35 = 0 - 0$

$18 x^{2} + 9 x - 35 = 0$ $18x^2 + 9x - 35 = 0$

How do you write 21x^2+5x-35=3x^2-4x21x2+5x−35=3x2−4x21x^2+5x-35=3x^2-4x in standard form?