That doesn't really scan. I'll guess that you have f(x) = 3x² + 6x - 3a, where x is your independent variable and a is some arbitrary constant. I'll also guess that you want to find f'(x) when x=-2. If so, read on....
Since the definition of the derivative ("first principles") is:
f'(x) = lim/h->0 [f(x+h) - f(x)] / h
...the derivative wanted is:
f'(2) = lim/x->0 [f(2 + h) - f(2)] / h
= lim/h->0 [(3(2 + h)² + 6(2 + h) - 3a) - (3(2)² + 6(2) - 3a)] / h
= lim/h->0 [3(4 + 4h + h²) + 6(2 + h) - 3a) - (3(4) + 6(2) - 3a]
The 3(4), 6(2) and -3a terms cancel in the numerator, leaving:
f'(2) = lim/h->0 [12h + 3h² + 6h] / h
= lim/h->0 (12 + 3h + 6)
= 12 + 6
= 18
Test that by taking the formal derivative of the polynomial f'(x) = 6x + 6, to see that f'(2) = 18 agrees.
Since the definition of the derivative ("first principles") is:
f'(x) = lim/h->0 [f(x+h) - f(x)] / h
...the derivative wanted is:
f'(2) = lim/x->0 [f(2 + h) - f(2)] / h
= lim/h->0 [(3(2 + h)² + 6(2 + h) - 3a) - (3(2)² + 6(2) - 3a)] / h
= lim/h->0 [3(4 + 4h + h²) + 6(2 + h) - 3a) - (3(4) + 6(2) - 3a]
The 3(4), 6(2) and -3a terms cancel in the numerator, leaving:
f'(2) = lim/h->0 [12h + 3h² + 6h] / h
= lim/h->0 (12 + 3h + 6)
= 12 + 6
= 18
Test that by taking the formal derivative of the polynomial f'(x) = 6x + 6, to see that f'(2) = 18 agrees.
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3x^2+7x+(2-3a)=0 , and use the quadratic formula to solve