Please note that there may be other methods apart from this large formulas to solve cubics and quartics. But I wanted to show here that the formulas do exist.

The general form of the **3rd degree**
equation (or
**Cubic**) is: ** ax ^{3}**

**Cubics** have **3** roots.

The **3 roots** can be represented this way:

**First root (of three):**

**Second root (of three):**

**Third root (of three):**

The second and third formula are equal except for a "+ or -" sign at
the beginning, and another "+ or -" sign in the middle. Note that
the second and third formula contain the imaginary unit "**i**".

Now, the same three formulas in **ASCII**. The differences
between the second and third formula are highlighted here in yellow:

x = -b/(3*a) -
(2^(1/3)*(-b^2 + 3*a*c))/(3*a*(-2*b^3 + 9*a*b*c - 27*a^2*d +
Sqrt[4*(-b^2 + 3*a*c)^3 + (-2*b^3 + 9*a*b*c - 27*a^2*d)^2])^(1/3)) +
(-2*b^3 + 9*a*b*c - 27*a^2*d + Sqrt[4*(-b^2 + 3*a*c)^3 + (-2*b^3 +
9*a*b*c - 27*a^2*d)^2])^(1/3)/(3*2^(1/3)*a)
x = -b/(3*a) + ((1 + i*Sqrt[3])*(-b^2 + 3*a*c))/(3*2^(2/3)*a*(-2*b^3 + 9*a*b*c - 27*a^2*d + Sqrt[4*(-b^2 + 3*a*c)^3 + (-2*b^3 + 9*a*b*c - 27*a^2*d)^2])^(1/3)) - (1 - i*Sqrt[3])*(-2*b^3 + 9*a*b*c - 27*a^2*d + Sqrt[4*(-b^2 + 3*a*c)^3 + (-2*b^3 + 9*a*b*c - 27*a^2*d)^2])^(1/3)/(6*2^(1/3)*a) x = -b/(3*a) + ((1 - i*Sqrt[3])*(-b^2 + 3*a*c))/(3*2^(2/3)*a*(-2*b^3 + 9*a*b*c - 27*a^2*d + Sqrt[4*(-b^2 + 3*a*c)^3 + (-2*b^3 + 9*a*b*c - 27*a^2*d)^2])^(1/3)) - (1 + i*Sqrt[3])*(-2*b^3 + 9*a*b*c - 27*a^2*d + Sqrt[4*(-b^2 + 3*a*c)^3 + (-2*b^3 + 9*a*b*c - 27*a^2*d)^2])^(1/3)/(6*2^(1/3)*a) |

The formulas on top were obtained with
Wolfram's program **Mathematica**. There seemed not to be an
option, so I copied them by hand into ASCII.

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