Plane cusp of type (2,5) ($x^2-y^5=0$)

This irreducible plane branch has parametrization $x=t^5$, $y=t^2$. Its only singular point is the origin. The resolution of the (2,5) cusp by two blowups makes its strict transform smooth in every characteristic.

Blow up the origin and use the chart $x=uy$, giving the strict transform $u^2-y^3=0$. Blow up its remaining singular point and use $u=vy$, giving $v^2-y=0$, which is smooth because its derivative in $y$ is $-1$. The complementary chart at each stage has no point of the strict transform on its exceptional divisor. These two blowups of a smooth algebraic surface resolve the branch, although the smooth transform is still tangent to an exceptional divisor.