The formula for solving an equation of second degree

can be derived like this:

If x² + px + q = 0, let x = y + m

then (y + m)² + p(y+m) + q = 0  and  y² + 2ym +m² + py +pm +q = 0
which leads to y² + 2y(2m + p) +m² +pm +q = 0 
If now m is chosen such that (2m +p) =  0  ie  m= -p/₂, the equation can be 
simplified to     y²+m² +pm +q = 0   or y²  = -m² -pm -q   and if    
m= -p/₂ is inserted in   the equation the result will be 
y²  = -(p²)/₄ +(p²)/₂ -q  = (p²)/₄ - q  and so:  
y  =   ±[(p²)/₄ - q] and as     x = y+m  = y + -p/₂,  the well known

formula:           x = -p/₂ ±[(p²)/₄ - q] immediately follows.

For a cubic equation

like x³ + px = q  the solution  can be achieved in a similar way,

In modern notation dal Ferros solution of x³ + px = q looks like this:

Say x = (y - b) and take the cubic of both sides:

(y - b) ³ = y³ - 3y²b +3yb² - b³
rearrange: (y-b)³ + 3yb(y-b) = y³- b³ ,
identify and let 3yb = p
which gives b = p/_3y and y³- b³ = q ,

then x=(y-b) will be a solution to x³ +px =q.

If now b = p/_3y is inserted in to y³- b³ = q the result will be y³- p³/_{27 y}3 = q,
which now gives us y⁶ -qy³ - p³/₂₇ =0, where we can get y³ from this
equation of second degree (in y³) and we also get b as b = p/_3y

and finally: x = ³[(p³/27 + q²/4) + q/2] - ³ [(p³/27 + q²/4) - q/2]

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