Solve x^2+2bx+4=0

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

Solve for b

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

$$x\neq 0$$

Steps for Solving Linear Equation

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

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

Subtract 4 from both sides.

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

The equation is in standard form.

$$2xb=-x^{2}-4$$

Divide both sides by 2x.

$$\frac{2xb}{2x}=\frac{-x^{2}-4}{2x}$$

Dividing by 2x undoes the multiplication by 2x.

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

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

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

Solve for x (complex solution)

$$x=\sqrt{b^{2}-4}-b$$

$$x=-\sqrt{b^{2}-4}-b$$

Solve for x

$$x=\sqrt{b^{2}-4}-b$$

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