Solve the equation with radicals $y=\left(\sqrt{x}\right)^{2x}$

$\sqrt{3}z+4=-\sqrt{3}\cdot z+8$

$\sqrt{\frac{\left(11\left(-1\right)\cdot x\right)^2}{2}+6x}=0$

$\sqrt{2}\left(x-3\right)+3\sqrt{2}\cdot\left(x+1\right)=8\sqrt{2}$

$\frac{1}{4}\cdot x+2\sqrt{x}+15=x$

$s\cdot \left(2\cdot 10+3\cdot 5-1\right)=\left(\sqrt{2}\right)^2+3^2$

$z\sqrt{3}+4=-\sqrt{3}\cdot z+8$

$\sqrt{z}-8=1$