can have solution of $x^4-3x^3+2x^2-3x+1=0$ using only high school methods can have solution of $x^4-3x^3+2x^2-3x+1=0$ using only high school methods???
i only know quadratic formula $$x =  \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ i tried many algebraic manipulations and i get $(x^2+1)^2=3(x^3+x)$, so can we have solution using only high school methods? i guess i have to do an algebraic substitution to reduce $x^3+x$ to polynomial of degree 2?? i also know $$(r_1 - r_2)^2 = (r_1 + r_2)^2 - 4r_1r_2$$ where $r_1,r_2$ are roots of polynomial of degree 2

Hint: $$\color{#C00}{x^4}\color{darkmagenta}{-3x^3}\color{royalblue}{+2x^2}\color{darkmagenta}{-3x}\color{#C00}{+1},\tag1$$ is a symmetric polynomial of degree $4$. I will describe here the basic procedure to solve such polynomial equations. Plugging in $x=0$ yields $1$, thus $0$ isn't a root and therefore $x\neq0$ which means that $x^2\neq0$. Thus it is legal to divide both sides by $x^2$, we should get $$x^2-3x+2-\dfrac3x+\dfrac1{x^2}=0\implies \left(x^2+\dfrac1{x^2}\right)-3\left(x+\dfrac1x\right)+2=0.\tag2$$ Now note that $$\left(x+\dfrac1x\right)^2=\left(x^2+\dfrac1{x^2}\right)+2\implies\left(x^2+\dfrac1{x^2}\right)=\left(x+\dfrac1x\right)^2-2.\tag3$$ From $(2)$ and $(3)$ we get $$\left(x+\dfrac1x\right)^2-3\left(x+\dfrac1x\right)=0.\tag4$$ This certainly looks like a quadratic, to convince yourself of that you can use the substitution $u\leadsto\left(x+\tfrac1x\right)$, and thus $(4)$ takes the usual form $$u^2-3u=0.$$ I'm sure you can finish it off from here.
