[latex]( \frac{ \sqrt[4]{2} }{ \sqrt{3} })^{x^2+4}= 20,25^{x+1} [/latex] Полное решение, пожалуйста

[latex](\frac{\sqrt[4]{2}}{\sqrt3})^{x^2+4}=20,25^{x+1}\\\\20,25=20\frac{1}{4}=\frac{81}{4}=\frac{3^4}{2^2}=\frac{(\sqrt3)^8}{(\sqrt[4]{2})^8}=(\frac{\sqrt3}{\sqrt[4]{2}})^8=(\frac{\sqrt[4]{2}}{\sqrt{3}})^{-8}\\\\(\frac{\sqrt[4]{2}}{\sqrt3})^{x^2+4}=(\frac{\sqrt[4]{2}}{\sqrt3})^{-8(x+1)}\\\\x^2+4=-8x-8\\\\x^2+8x+12=0\\\\x_1=-6,\; x_2=-2\; (teor.\; Vieta)[/latex]