求解 {l}{yleq-2x^2+12x-6}{ygeqx^2+x-12}

$$\left\{ \begin{array} { l } { y \leq - 2 x ^ { 2 } + 12 x - 6 } \\ { y \geq x ^ { 2 } + x - 12 } \end{array} \right.$$

求解 x 的值

$$\left\{\begin{matrix}x=3\text{, }&y\geq 0\text{ and }y\leq 12\\x=\frac{\sqrt{4y+49}-1}{2}\text{, }&y=\frac{-7\sqrt{193}-13}{9}\\x\in \begin{bmatrix}\frac{-\sqrt{4y+49}-1}{2},\frac{\sqrt{24-2y}}{2}+3\end{bmatrix}\text{, }&-\frac{\sqrt{96-8y}}{4}+3<\frac{-\sqrt{4y+49}-1}{2}\text{ and }\frac{\sqrt{96-8y}}{4}+3\leq \frac{\sqrt{4y+49}-1}{2}\text{ and }y<12\text{ and }y>-\frac{49}{4}\\x\in \begin{bmatrix}-\frac{\sqrt{24-2y}}{2}+3,\frac{\sqrt{24-2y}}{2}+3\end{bmatrix}\text{, }&-\frac{\sqrt{96-8y}}{4}+3\geq \frac{-\sqrt{4y+49}-1}{2}\text{ and }\frac{\sqrt{96-8y}}{4}+3\leq \frac{\sqrt{4y+49}-1}{2}\text{ and }y<12\text{ and }y>-\frac{49}{4}\\x\in \begin{bmatrix}\frac{-\sqrt{4y+49}-1}{2},\frac{\sqrt{4y+49}-1}{2}\end{bmatrix}\text{, }&\frac{\sqrt{4y+49}-1}{2}<\frac{\sqrt{96-8y}}{4}+3\text{ and }-\frac{\sqrt{96-8y}}{4}+3<\frac{-\sqrt{4y+49}-1}{2}\text{ and }y>-\frac{49}{4}\text{ and }y<12\\x\in \begin{bmatrix}-\frac{\sqrt{24-2y}}{2}+3,\frac{\sqrt{4y+49}-1}{2}\end{bmatrix}\text{, }&-\frac{\sqrt{96-8y}}{4}+3\geq \frac{-\sqrt{4y+49}-1}{2}\text{ and }-\frac{\sqrt{96-8y}}{4}+3<\frac{\sqrt{4y+49}-1}{2}\text{ and }\frac{\sqrt{4y+49}-1}{2}<\frac{\sqrt{96-8y}}{4}+3\text{ and }y<12\text{ and }y>-\frac{49}{4}\end{matrix}\right.$$

求解 y 的值

$$y\in \begin{bmatrix}x^{2}+x-12,-2x^{2}+12x-6\end{bmatrix}\text{, }x\geq \frac{11-\sqrt{193}}{6}\text{ and }x\leq \frac{\sqrt{193}+11}{6}$$