Quadratic Functions Question

$f(0)=2 \Rightarrow c=2 \\ \\ \text{Then we have }a+b+2=-1 \Rightarrow b=-3-a$

$\text{ so we have }f(x) = a x^2 -(3+a)x + 2$

$\text{This has zeros at }\\ x = \dfrac{(3+a)\pm \sqrt{(3+a)^2-8a}}{2a}$

$\text{subtracting we have}\\ \dfrac{\sqrt{(3+a)^2-8a}}{a}=\dfrac{\sqrt{9-2a+a^2}}{a} = 2\sqrt{2}$

$\dfrac{9-2a+a^2}{a^2}= 8 \\ \\ 7a^2+2a-9=0 \\ \\ a=\dfrac{-2\pm \sqrt{4+252}}{14}=\dfrac{-2\pm 16}{14} = 1,~-\dfrac 9 7$

$f(x) = x^2 -4x+2 \text{ or }-\dfrac 9 7 x^2 - \dfrac{12}{7}+2$