What is the answer to solvefor a,n^2-n=a(n-4)^2+b(n-4)+c ?

The answer to solvefor a,n^2-n=a(n-4)^2+b(n-4)+c is a=(n^2-n-b(n-4)-c)/((-4+n)^2);\quad n\ne 4

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solvefor a,n^2-n=a(n-4)^2+b(n-4)+c

solve for

a, n^{2} - n = a (n - 4)^{2} + b (n - 4) + c

a = \frac{n ^{2} - n - b ( n - 4 ) - c}{( - 4 + n ) ^{2}}; n \neq = 4

Solution steps

n^{2} - n = a (n - 4)^{2} + b (n - 4) + c

Switch sides

a (n - 4)^{2} + b (n - 4) + c = n^{2} - n

Move

b (n - 4)

to the right side

a (n - 4)^{2} + c = n^{2} - n - b (n - 4)

Move

c

to the right side

a (n - 4)^{2} = n^{2} - n - b (n - 4) - c

Divide both sides by

(n - 4)^{2}; n \neq = 4

a = \frac{n ^{2} - n - b ( n - 4 ) - c}{( - 4 + n ) ^{2}}; n \neq = 4

x, y - 2 x + 3 = 0

x = 100000 \cdot e^{0.06 \cdot 20}

A, y = A x^{2} + B e^{x}

f = lo g_{\frac{1}{4}} (14 + 2)

4 (x - 2) = - 8 (x - (- 4))

What is the answer to solvefor a,n^2-n=a(n-4)^2+b(n-4)+c ?
The answer to solvefor a,n^2-n=a(n-4)^2+b(n-4)+c is a=(n^2-n-b(n-4)-c)/((-4+n)^2);\quad n\ne 4