Solve 0=x^2+bx+3

$$0 = x ^ { 2 } + b x + 3$$

Solve for b

$$b=-x-\frac{3}{x}$$

$$x\neq 0$$

Steps for Solving Linear Equation

Swap sides so that all variable terms are on the left hand side.

$$x^{2}+bx+3=0$$

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

$$bx+3=-x^{2}$$

Subtract 3 from both sides.

$$bx=-x^{2}-3$$

The equation is in standard form.

$$xb=-x^{2}-3$$

Divide both sides by x.

$$\frac{xb}{x}=\frac{-x^{2}-3}{x}$$

Dividing by x undoes the multiplication by x.

$$b=\frac{-x^{2}-3}{x}$$

Divide -x^{2}-3 by x.

$$b=-x-\frac{3}{x}$$

Solve for x (complex solution)

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

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

Solve for x

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

$$x=\frac{-\sqrt{b^{2}-12}-b}{2}\text{, }|b|\geq 2\sqrt{3}$$