Solve 2^3-kz^2+21z-20=0

$$2 ^ { 3 } - k z ^ { 2 } + 21 z - 20 = 0$$

Solve for k

$$k=\frac{3\left(7z-4\right)}{z^{2}}$$

$$z\neq 0$$

Steps for Solving Linear Equation

Calculate 2 to the power of 3 and get 8.

$$8-kz^{2}+21z-20=0$$

Subtract 21z from both sides. Anything subtracted from zero gives its negation.

$$8-kz^{2}-20=-21z$$

Add 20 to both sides.

$$8-kz^{2}=-21z+20$$

Subtract 8 from both sides.

$$-kz^{2}=-21z+20-8$$

Subtract 8 from 20 to get 12.

$$-kz^{2}=-21z+12$$

The equation is in standard form.

$$\left(-z^{2}\right)k=12-21z$$

Divide both sides by -z^{2}.

$$\frac{\left(-z^{2}\right)k}{-z^{2}}=\frac{12-21z}{-z^{2}}$$

Dividing by -z^{2} undoes the multiplication by -z^{2}.

$$k=\frac{12-21z}{-z^{2}}$$

Divide -21z+12 by -z^{2}.

$$k=-\frac{3\left(4-7z\right)}{z^{2}}$$

Solve for z

$$\left\{\begin{matrix}z=-\frac{\sqrt{3\left(147-16k\right)}-21}{2k}\text{; }z=\frac{\sqrt{3\left(147-16k\right)}+21}{2k}\text{, }&k\neq 0\text{ and }k\leq \frac{147}{16}\\z=\frac{4}{7}\text{, }&k=0\end{matrix}\right.$$