(a) "f(1)=7" and "f(2)=10" give "a+b-2=7" and "4a+2b-2=10". By the first equation, "b=9-a", so substituting this to the second one gives
"4a+2(9\u2212a)=12\u21d2a+9=6\u21d2a=\u22123,b=12".
(b) We are to solve the quadratic equation
"-3x^2+12x-2=3x+1\\Leftrightarrow 0=3x^2-9x+3\\Rightarrow x=\\frac{3\\pm\\sqrt5}{2}"
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