The normal to the curve $5x^5-10x^3+x+2y+6=0$ at $A(0,-3)$ meets the curve again at two points $B$ and $C$. The length of the line segment $BC$ is:

Given the curve: $5x^5-10x^3+x+2y+6=0$.Differentiating with respect to $x$:$25x^4-30x^2+1+2\frac{dy}{dx}=0⟹\frac{dy}{dx}=\frac{-(25x^4-30x^2+1)}{2}$ At $A(0,-3)$, the slope of the tangent is:$m_t=\frac{-(0-0+1)}{2}=-\frac{1}{2}$ The slope of the normal ($m_n$) is the negative reciprocal: $m_n=2$.Equation of the normal at $A(0,-3)$:$y-(-3)=2(x-0)⟹y=2x-3$ Substituting $y=2x-3$ into the curve equation:$5x^5-10x^3+x+2(2x-3)+6=0$$5x^5-10x^3+x+4x-6+6=0$$5x^5-10x^3+5x=0$$5x(x^4-2x^2+1)=0⟹5x(x^2-1{)}^2=0$$x=0$ (Point $A$), and $x^2=1⟹x=1,-1$ (Points $B$ and $C$).For $x=1$, $y=2(1)-3=-1$. Point $B$ is $(1,-1)$.For $x=-1$, $y=2(-1)-3=-5$. Point $C$ is $(-1,-5)$.Length $BC=\sqrt{(-1-1{)}^2+(-5-(-1){)}^2}=\sqrt{(-2{)}^2+(-4{)}^2}=\sqrt{4+16}=\sqrt{20}=2\sqrt{5}$.