How do you find c in $(4 x^{2}) - 12 x + c = 0$ $(4x^2)-12x+c=0$ ?

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Answer 1 · 2018-06-13T13:18:44

$c = 4 x (3 - x)$ $c= 4x(3-x)$

$4 x^{2} - 12 x + c = 0$ $4x^2-12x+c=0$

$c = - (4 x^{2} - 12 x)$ $c=-(4x^2-12x)$

$= - 4 x (x - 3)$ $= -4x(x-3)$

$= 4 x (3 - x)$ $= 4x(3-x)$

How do you find c in (4x2)−12x+c=0(4x^2)-12x+c=0 ?