finding a tangent line to a parabola

Hint: $$x^2+5x+3=x+b\iff x^2+4x+3-b=0,$$ this quadratic equation should have exactly one solution therefore the discriminant $\Delta=?$

Note that you like the line $y=x+b$ to be tangent to the parabola. The slope of this line is $ m=1.$ Thus the derivative of your parabola should be the same as the slope of the tangent line. Find the point of the tangency and find the $b$ value so that the line passes through that point.

The tangent to parabola has slope $y' = 2x+5$ which should be slope of line =1 at the point of contact. So point of contact is at x=-2. Substituting in parabola,we get y=-3. Now this point lies on line as well.