Question 870317
It's not a quadratic, it's a cubic.
Try the factors of 9 first, 1, 3 & 9 plus and minus
If that doesn't work, try the factors /2, then /4
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You can use an Excel sheet to do the arithmetic, or a graphing calculator, or graphical methods.
If you find 1 real zero, then a quadratic will be left.
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f(x) = 4x^3 - 10x^2 + 9
f'(x) = 12x^2 - 20x = 0
x = 0, x = 5/3
f(0) = 9, f(5/3) <0
--> a zero between 0 and 5/3