Solve y=(2x^2+x)^200

$$y = ( 2 x ^ { 2 } + x ) ^ { 200 }$$

Solve for x

$$\left\{\begin{matrix}x=\frac{-\sqrt{8\sqrt[200]{y}+1}-1}{4}\text{; }x=\frac{\sqrt{8\sqrt[200]{y}+1}-1}{4}\text{, }&y\geq 0\\x=\frac{-\sqrt{-8\sqrt[200]{y}+1}-1}{4}\text{; }x=\frac{\sqrt{-8\sqrt[200]{y}+1}-1}{4}\text{, }&y\geq 0\text{ and }y\leq \frac{1}{4149515568880992958512407863691161151012446232242436899995657329690652811412908146399707048947103794288197886611300789182395151075411775307886874834113963687061181803401509523685376}\end{matrix}\right.$$

Solve for y

$$y=\left(x\left(2x+1\right)\right)^{200}$$