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Two successive point blowups of a smooth algebraic surface suffice. Let $X_2\to X_1={\mathbb{A}}^2$ be the blowup of the affine plane at the origin. In its chart $x=uy$, with coordinates $(y,u)$, the total-transform equation isRemoving the exceptional factor gives the strict transformAbove the original origin it has just one point, $(y,u)=(0,0)$, which is still singular. The other chart is $y=vx$, where the strict transform has equation $1-v^5{x}^3=0$ and does not meet the exceptional divisor $x=0$. Thus there are no other points above the origin to resolve.

Blow up the remaining point to obtain $X_3\to X_2$. In the chart $u=vy$, with coordinates $(y,v)$, the total transform of $C_2$ isand hence its strict transform isThe derivative of $v^2-y$ with respect to $y$ is $-1$, so this is nonsingular, even in characteristics two or five. In the other chart $y=wu$, the strict transform has equation $1-w^{3}u=0$ and does not meet the exceptional divisor $u=0$. Therefore the only point of $C_3$ mapping to the original origin is the smooth point $(y,v)=(0,0)$.

The required sequence isThe local parameter $v$ there gives $y=v^2$ and $x=v^5$, also verifying the resolved branch directly. This is the resolution of the (2,5) cusp by two blowups. Its tangency to an exceptional divisor does not affect the requested nonsingularity of the strict transform; making the whole total transform have normal crossings is a stronger task.

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.