Find a solution x=x(t) of the equation x'+2x=t^{2}+4t+7 in the form of a quadratic function of t, that is, of the form x(t) = at^{2}+bt +c, where a,b, and c are to be determined.

At first I thought you would solve this by finding the integrating factor, but after speaking with the teacher he said to plug in the form x(t)=at^{2}+bt+c into the Diff. Eq. and use the resulting expression to solve for a,b, and c by comparing the expression on the left side to the expression on the right side.

So now I get this, but I am not sure what to do from here.

x'(t)=2t+4

2t^{2}+10t+18 = t^{2}+4t+7