Помогите решить пожалуйстa

$\tt (x^2+4)^2+5(x^2+4)-24=0\\x^2+4=y\\y^2+5y-24=0\\D=b^2-4ac=25+4\cdot 24=121=11^2\\y_{1,2}=\frac{-b\pm\sqrt{D}}{2a}=\frac{-5\pm11}{2}=\left |{ {{3} \atop {-8}} \right. \\\left [{ {{x^2+4=3} \atop {x^2+4=-8}} \right. \left [{ {{x^2\neq -1} \atop {x^2\neq -12}} \right.$

Ответ: корней нет

$\tt (x^2+x+2)(x^2+x-5)=-10\\x^2+x+2=z\\z(z-7)+10=0\\z^2-7z+10=0\\D=49-40=9=3^2\\z_{1,2}=\frac{7\pm 3}{2}=\left |{ {{5} \atop {2}} \right. \\\left [{ {{x^2+x+2=5} \atop {x^2+x+2=2}} \right.$

$\tt x^2+x+2=5\\x^2+x-3=0\\D=1+12=13=(\sqrt{13})^2\\x_{1,2}=\frac{-1\pm\sqrt{13}}{2}$

$\tt x^2+x+2=2\\x^2+x=0\\x(x+1)=0\\\left [{ {{x=-1} \atop {x=0}} \right.$

Ответ: $\tt \frac{-1-\sqrt{13}}{2};\frac{-1+\sqrt{13}}{2};-1;0$