Answer
$\left\{x\ |\ x\geq-1 \right\}$ or $\left[-1,+\infty \right)$
Work Step by Step
$x(4x+3)\leq(2x+1)^{2}$
$4x^{2}+3x \leq 4x^{2}+4x+1 \qquad$ ... add $ -4x-4x^{2}$
$-x \leq 1\qquad \qquad$... multiply with $-1 $ (negative)
... the inequality changes direction ...
$ x\geq-1$
Solution set: $\left\{x\ |\ x\geq-1 \right\}$ or $\left[-1,+\infty \right)$
