Solve for: -3x^2-4x+4 <= 0

Expression: $-3{x}^{2}-4x+4 \leq 0$

Write $-4x$ as a difference

$-3{x}^{2}+2x-6x+4 \leq 0$

Factor out $-x$ from the expression

$-x \times \left( 3x-2 \right)-6x+4 \leq 0$

Factor out $-2$ from the expression

$-x \times \left( 3x-2 \right)-2\left( 3x-2 \right) \leq 0$

Factor out $-\left( 3x-2 \right)$ from the expression

$-\left( 3x-2 \right) \times \left( x+2 \right) \leq 0$

Change the signs on both sides of the inequality and flip the inequality sign

$\left( 3x-2 \right) \times \left( x+2 \right) \geq 0$

Separate the inequality into two possible cases

$\begin{array} { l }\left\{\begin{array} { l } 3x-2 \geq 0 \\ x+2 \geq 0\end{array} \right.,\\\left\{\begin{array} { l } 3x-2 \leq 0 \\ x+2 \leq 0\end{array} \right.\end{array}$

Solve the inequality for $x$

$\begin{array} { l }\left\{\begin{array} { l } x \geq \frac{ 2 }{ 3 } \\ x+2 \geq 0\end{array} \right.,\\\left\{\begin{array} { l } 3x-2 \leq 0 \\ x+2 \leq 0\end{array} \right.\end{array}$

Solve the inequality for $x$

$\begin{array} { l }\left\{\begin{array} { l } x \geq \frac{ 2 }{ 3 } \\ x \geq -2\end{array} \right.,\\\left\{\begin{array} { l } 3x-2 \leq 0 \\ x+2 \leq 0\end{array} \right.\end{array}$

Solve the inequality for $x$

$\begin{array} { l }\left\{\begin{array} { l } x \geq \frac{ 2 }{ 3 } \\ x \geq -2\end{array} \right.,\\\left\{\begin{array} { l } x \leq \frac{ 2 }{ 3 } \\ x+2 \leq 0\end{array} \right.\end{array}$

Solve the inequality for $x$

$\begin{array} { l }\left\{\begin{array} { l } x \geq \frac{ 2 }{ 3 } \\ x \geq -2\end{array} \right.,\\\left\{\begin{array} { l } x \leq \frac{ 2 }{ 3 } \\ x \leq -2\end{array} \right.\end{array}$

Find the intersection

$\begin{array} { l }x \in \left[ \frac{ 2 }{ 3 }, +\infty\right\rangle,\\\left\{\begin{array} { l } x \leq \frac{ 2 }{ 3 } \\ x \leq -2\end{array} \right.\end{array}$

Find the intersection

$\begin{array} { l }x \in \left[ \frac{ 2 }{ 3 }, +\infty\right\rangle,\\x \in \left\langle-\infty, -2\right]\end{array}$

Find the union

$x \in \left\langle-\infty, -2\right] \cup \left[ \frac{ 2 }{ 3 }, +\infty\right\rangle$