$a)$

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

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

$x{\left(x-2\right)}^2=0$

$x=0\vee x=2$

$\overline{}$

$b)$

$2x^4+5x^3-3x^2=0$

$x^2\left(2x^2+5x-3\right)=0$

$x=0\vee 2x^2+5x-3=0$

$\Delta =5^2-4\cdot 2\cdot \left(-3\right)=25+24=49$

$\sqrt{\Delta }=7$