Quadratic Identity question

QUESTION:
Write $3x^2+7$ in the form $a(x-2)^2+b(x+3)+c$.


ANSWER:
$3x^2+7=a(x-2)^2+b(x+3)+c$ holds for all $x$.

Best way is to just expand the RHS and then equate coefficients of the powers of $x$ on LHS and RHS.

$3x^2+7=a(x^2-4x+4)+b(x+3)+c$
$3x^2+7=a x^2-4ax+4a+bx+3b+c$
$3x^2+7=a x^2+(b-4a)x+(4a+3b+c)$

Therefore 

1. $a=3$  (comparing coefficient of $x^2$ on LHS and RHS)
2. $b-4a=0$ since coefft of $x$ on LHS is 0. (there is no $x$ term)
3. $4a+3b+c=7$ (Constant terms on LHS and RHS)

So, $b-4(3)=0$.

$b=12$.

Finally, $4(3)+3(12)+c=7$.

So, $c=7-12-36=-41$.

Final answer: $a=3, b=12, c=-41$.