Answer
$${\text{Continuous everywhere}}{\text{.}}$$
Work Step by Step
$$\eqalign{
& {\text{Let the function }}f\left( x \right) = \frac{{2x + 1}}{{4{x^2} + 4x + 5}} \cr
& {\text{This is a rational function}}{\text{, and a rational function is continuous }} \cr
& {\text{at every point where the denominator is nonzero}}{\text{, and has }} \cr
& {\text{discontinuities at the points where the denominator is zero}}{\text{.}} \cr
& {\text{Setting the denominator to 0}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4{x^2} + 4x + 5 = 0 \cr
& {\text{Solving the equation by the quadratic formula}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = \frac{{ - \left( { - 4} \right) \pm \sqrt {{{\left( 4 \right)}^2} - 4\left( 4 \right)\left( 5 \right)} }}{{2\left( 4 \right)}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = \frac{{4 \pm \sqrt { - 64} }}{8} \cr
& {\text{Thus}}{\text{, there are no real values at which the denominator is zero}} \cr
& {\text{then}}{\text{, the function is continuous for all real numbers}}{\text{.}} \cr
& \cr
& {\text{Continuous everywhere}}{\text{.}} \cr} $$
