Just like if we have any quadratic equation which has complex roots, then we are not able to factorize it easily. So we apply quadratic formula and get the roots.

Similarly if we have a cubic equation which has two complex roots (which we know conjugate of each other) and one fractional root, then we are not able to find its first root by hit & trial.

So my question is like quadratic formula, is there exist any thing like cubic formula which help in solving cubic equations?

For example, I have an equation $$2x^3+9x^2+9x-7=0\tag{1}$$ and I have to find its solution which I am not able to find because it has no integral solution. Its solutions are $\dfrac {1}{2}$, $\dfrac{-5\pm \sqrt{3}i}{2} $, I know these solutions because this equation is generated by myself.

So how can I solve equations like these?

Also while typing this question, I thought about the derivation of quadratic formula, which is derived by completing the square method.

So I tried to apply ‘completing the cube’ method on the general equation $ax^3+bx^2+cx+d=0$ but it didn't help.

So please help me in finding a cubic formula or to solve the equations like given in example by an alternative method.

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    $\begingroup$ see here mathworld.wolfram.com/CubicFormula.html $\endgroup$ Dec 10, 2016 at 15:59
  • $\begingroup$ When there is only one real root, there is Cardano's formula (which is actually valid in all cases, but has to be interpreted). $\endgroup$
    – Bernard
    Dec 10, 2016 at 16:00
  • 1
    $\begingroup$ For cubics with rational coefficients, you can always multiply through by the LCM of the denominators to get something with integer coeffs, and then apply the rational roots theorem, which would have helped you find $\frac{1}{2}$ in your problem. $\endgroup$ Dec 10, 2016 at 16:03
  • 3
    $\begingroup$ Google "cubic equation wikipedia" and "quartic equation wikipedia" $\endgroup$ Dec 10, 2016 at 16:17
  • $\begingroup$ If (after a while) you get interested in the solution of "quintic equation" (an equation of the form $ax^5+bx^4+cx^3+dx^2+ex+f=0$ see this. $\endgroup$
    – user170039
    Dec 11, 2016 at 4:09

3 Answers 3


Yes, we do have a cubic formula! By Cardan's Method...

Cardan's Method: To solve the general cubic$$x^3+ax^2+bx+c=0\tag{i}$$ Remove the $ax^2$ term by substituting $x=\dfrac {y-a}3$. Let the transformed equation be$$y^3+qy+r=0\tag{ii}$$ To solve this depressed cubic, substitute $y=u+v$ to get$$u^3+v^3+(3uv+q)(u+v)+r=0\tag{iii}$$ Put $3uv+q=0$ to get $u=-\dfrac q{3v}$ and substituting this back gives a quadratic in $v^3$. The roots of the quadratic are equal to $u^3,v^3$ respectively. And from our substitution, we get a root as$$y=\left\{-\frac r2+\sqrt{\frac {r^2}4+\frac {q^3}{27}}\right\}^{\frac 13}+\left\{-\frac r2-\sqrt{\frac {r^2}4+\frac {q^3}{27}}\right\}^{\frac 13}\tag{iv}$$ With the other two roots found with the cube roots of unity.

To find the original root of $(\text i)$, substitute $y$ into your transformation.

  • 1
    $\begingroup$ “With the other two roots found with the cube roots of unity.” Explain please $\endgroup$ Jul 13, 2020 at 13:48
  • $\begingroup$ @RadialArmSaw, let w = 1/2+sqrt(3)/2*i. Since you already found the one multiplied by 1, I think you have to multiply your solution by w and w^2 if I'm correct. $\endgroup$
    – jiggzson
    Mar 20, 2021 at 23:39
  • $\begingroup$ The general formula is $$x=-\frac{b}{3a}+w_1\sqrt[3]{\frac{9abc-2b^3-27a^2d}{54a^3}+\sqrt{(\frac{9abc-2b^3-27a^2d}{54a^3})^2+(\frac{3ac-b^2}{9a^2})^3}}+w_2\sqrt[3]{\frac{9abc-2b^3-27a^2d}{54a^3}-\sqrt{(\frac{9abc-2b^3-27a^2d}{54a^3})^2+(\frac{3ac-b^2}{9a^2})^3}}$$ $\endgroup$ Feb 24 at 4:29
  • $\begingroup$ In the above formula, either $w_1=w_2=1$, $w_1=\frac{-1+i\sqrt3}{2}$ and $w_2=\frac{-1-i\sqrt3}{2}$, or $w_1=\frac{-1-i\sqrt3}{2}$ and $w_2=\frac{-1+i\sqrt3}{2}$ $\endgroup$ Feb 24 at 4:32

One method is to depress the cubic, then apply trig functions.


Divide both sides by $s$ to get:


Let $x=y-\frac a3$ to get


If $d>0$, then use the trigonometric identity:




We exploit this identity by letting $y=fz$ and multiplying both sides by $g$ to get


$$\begin{cases}4=f^3g\\3=dfg\end{cases}\implies\begin{cases}f=2\sqrt{\frac d3}\\g=\frac{3\sqrt3}{2d^{3/2}}\end{cases}$$



$$\implies z=-\sinh\left(\frac13\operatorname{arcsinh}\left(\frac{3e\sqrt3}{2d^{3/2}}\right)\right)$$

$$\implies x=-2\sqrt{\frac d3}\sinh\left(\frac13\operatorname{arcsinh}\left(\frac{3e\sqrt3}{2d^{3/2}}\right)\right)-\frac a3$$

If $d<0$, use $\cos(3\theta)$ or $\cosh(3\theta)$ and respective triple angle formulas.





, $i^2=-1$, and exactly one of the following is true.


or $w_1=\frac{-1+i\sqrt3}{2}$,and$w_2=\frac{-1-i\sqrt3}{2}$

or $w_1=\frac{-1-i\sqrt3}{2}$, and $w_2=\frac{-1+i\sqrt3}{2}$



Credit goes to Girolamo Cardano.

This formula can be easily memorized once you notice certain patterns.


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