$\text{a)}{x}^3-7x^2+12x=0$

$x\left(x^2-7x+12\right)=0$

$x=0\text{lub}{x}^2-7x+12=0$

$\Delta ={\left(-7\right)}^2-4\cdot 1\cdot 12=49-48=1,\sqrt{\Delta }=1$

$x=\frac{7-1}{2}=\frac{6}{2}=3\text{lub}x=\frac{7+1}{2}=\frac{8}{2}=4$

$\underline{x\in \left\{0,3,4\right\}}$

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$\text{b)}-2x^4+9x^3+5x^2=0$

$x^2\left(-2x^2+9x+5\right)=0$

$x=0\text{lub}-2x^2+9x+5=0$

$\Delta =9^2-4\cdot \left(-2\right)\cdot 5=81+40=121,\sqrt{\Delta }=11$

$x=\frac{-9-11}{2\cdot \left(-2\right)}=\frac{-20}{-4}=5\text{lub}x=\frac{-9+11}{2\cdot \left(-2\right)}=\frac{2}{-4}=-\frac{1}{2}$

$\underline{x\in \left\{-\frac{1}{2},0,5\right\}}$

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$\text{c)}3x^3+4x^2+x=0$

$x\left(3x^2+4x+1\right)=0$

$x=0\text{lub}3x^2+4x+1=0$

$\Delta =4^2-4\cdot 3\cdot 1=16-12=4,\sqrt{\Delta }=2$

$x=\frac{-4-2}{2\cdot 3}=\frac{-6}{6}=-1\text{lub}x=\frac{-4+2}{2\cdot 3}=\frac{-2}{6}=-\frac{1}{3}$

$\underline{x\in \left\{-1,-\frac{1}{3},0\right\}}$

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$\text{d)}4x^5-3x^4+2x^3=0$

$x^3\left(4x^2-3x+2\right)=0$

$x=0\text{lub}4x^2-3x+2=0$

$\Delta ={\left(-3\right)}^2-4\cdot 4\cdot 2<0$

Δ<0, więc równanie kwadratowe jest sprzeczne. Zatem jedynym rozwiązaniem początkowego równania jest:

$x=0$  

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$\text{e)}2x^2-3x=x-x^3$

$x^3+2x^2-4x=0$

$x\left(x^2+2x-4\right)=0$

$x=0\text{lub}{x}^2+2x-4=0$

$\Delta =2^2-4\cdot 1\cdot \left(-4\right)=4+16=20,\sqrt{\Delta }=2\sqrt{5}$

$x=\frac{-2-2\sqrt{5}}{2}=-1-\sqrt{5}\text{lub}x=\frac{-2+2\sqrt{5}}{2}=-1+\sqrt{5}$

$\underline{x\in \left\{-1-\sqrt{5},0,-1+\sqrt{5}\right\}}$

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$\text{f)}2x^6-x^5=x^5+x^4$  

$2x^6-2x^5-x^4=0$  

$x^4\left(2x^2-2x-1\right)=0$  

$x=0\text{lub}2x^2-2x-1=0$  

$\Delta ={\left(-2\right)}^2-4\cdot 2\cdot \left(-1\right)=4+8=12,\sqrt{\Delta }=2\sqrt{3}$  

$x=\frac{2-2\sqrt{3}}{4}=\frac{1-\sqrt{3}}{2}\text{lub}x=\frac{2+2\sqrt{3}}{4}=\frac{1+\sqrt{3}}{2}$  

$\underline{x\in \left\{\frac{1-\sqrt{3}}{2},0,\frac{1+\sqrt{3}}{2}\right\}}$