Solve 19a^3-12x+4=0

$$19 a ^ { 3 } - 12 x + 4 = 0$$

Solve for x

$$x=\frac{19a^{3}}{12}+\frac{1}{3}$$

Steps for Solving Linear Equation

Subtract 19a^{3} from both sides. Anything subtracted from zero gives its negation.

$$-12x+4=-19a^{3}$$

Subtract 4 from both sides.

$$-12x=-19a^{3}-4$$

Divide both sides by -12.

$$\frac{-12x}{-12}=\frac{-19a^{3}-4}{-12}$$

Dividing by -12 undoes the multiplication by -12.

$$x=\frac{-19a^{3}-4}{-12}$$

Divide -19a^{3}-4 by -12.

$$x=\frac{19a^{3}}{12}+\frac{1}{3}$$

Solve for a (complex solution)

$$a=\frac{19^{\frac{2}{3}}e^{\frac{2\pi i}{3}}\sqrt[3]{12x-4}}{19}$$

$$a=\frac{19^{\frac{2}{3}}\sqrt[3]{12x-4}}{19}$$

$$a=\frac{19^{\frac{2}{3}}e^{\frac{4\pi i}{3}}\sqrt[3]{12x-4}}{19}$$

Solve for a

$$a=\frac{19^{\frac{2}{3}}\sqrt[3]{12x-4}}{19}$$