Solve bx^2+cx+4=0

$$b x ^ { 2 } + c x + 4 = 0$$

Solve for b

$$b=-\frac{cx+4}{x^{2}}$$

$$x\neq 0$$

Steps for Solving Linear Equation

Subtract cx from both sides. Anything subtracted from zero gives its negation.

$$bx^{2}+4=-cx$$

Subtract 4 from both sides.

$$bx^{2}=-cx-4$$

The equation is in standard form.

$$x^{2}b=-cx-4$$

Divide both sides by x^{2}.

$$\frac{x^{2}b}{x^{2}}=\frac{-cx-4}{x^{2}}$$

Dividing by x^{2} undoes the multiplication by x^{2}.

$$b=\frac{-cx-4}{x^{2}}$$

Divide -cx-4 by x^{2}.

$$b=-\frac{cx+4}{x^{2}}$$

Solve for c

$$c=-bx-\frac{4}{x}$$

$$x\neq 0$$

Steps for Solving Linear Equation

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

$$cx+4=-bx^{2}$$

Subtract 4 from both sides.

$$cx=-bx^{2}-4$$

The equation is in standard form.

$$xc=-bx^{2}-4$$

Divide both sides by x.

$$\frac{xc}{x}=\frac{-bx^{2}-4}{x}$$

Dividing by x undoes the multiplication by x.

$$c=\frac{-bx^{2}-4}{x}$$

Divide -bx^{2}-4 by x.

$$c=-bx-\frac{4}{x}$$