Solve x^2+bx+12=0

$$x ^ { 2 } + b x + 12 = 0$$

Solve for b

$$b=-x-\frac{12}{x}$$

$$x\neq 0$$

Steps for Solving Linear Equation

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

$$bx+12=-x^{2}$$

Subtract 12 from both sides.

$$bx=-x^{2}-12$$

The equation is in standard form.

$$xb=-x^{2}-12$$

Divide both sides by x.

$$\frac{xb}{x}=\frac{-x^{2}-12}{x}$$

Dividing by x undoes the multiplication by x.

$$b=\frac{-x^{2}-12}{x}$$

Divide -x^{2}-12 by x.

$$b=-x-\frac{12}{x}$$

Solve for x (complex solution)

$$x=\frac{\sqrt{b^{2}-48}-b}{2}$$

$$x=\frac{-\sqrt{b^{2}-48}-b}{2}$$

Solve for x

$$x=\frac{\sqrt{b^{2}-48}-b}{2}$$

$$x=\frac{-\sqrt{b^{2}-48}-b}{2}\text{, }|b|\geq 4\sqrt{3}$$