Solve 21c^2-b4c+48=0

$$21 c ^ { 2 } - b 4 c + 48 = 0$$

Solve for b_4

$$b_{4}=21c+\frac{48}{c}$$

$$c\neq 0$$

Steps for Solving Linear Equation

Subtract 21c^{2} from both sides. Anything subtracted from zero gives its negation.

$$-b_{4}c+48=-21c^{2}$$

Subtract 48 from both sides.

$$-b_{4}c=-21c^{2}-48$$

The equation is in standard form.

$$\left(-c\right)b_{4}=-21c^{2}-48$$

Divide both sides by -c.

$$\frac{\left(-c\right)b_{4}}{-c}=\frac{-21c^{2}-48}{-c}$$

Dividing by -c undoes the multiplication by -c.

$$b_{4}=\frac{-21c^{2}-48}{-c}$$

Divide -21c^{2}-48 by -c.

$$b_{4}=21c+\frac{48}{c}$$

Solve for c (complex solution)

$$c=\frac{\sqrt{b_{4}^{2}-4032}+b_{4}}{42}$$

$$c=\frac{-\sqrt{b_{4}^{2}-4032}+b_{4}}{42}$$

Solve for c

$$c=\frac{\sqrt{b_{4}^{2}-4032}+b_{4}}{42}$$

$$c=\frac{-\sqrt{b_{4}^{2}-4032}+b_{4}}{42}\text{, }|b_{4}|\geq 24\sqrt{7}$$