Solve 2x^2+kx+5=0

$$2 x ^ { 2 } + k x + 5 = 0$$

Solve for k

$$k=-2x-\frac{5}{x}$$

$$x\neq 0$$

Steps for Solving Linear Equation

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

$$kx+5=-2x^{2}$$

Subtract 5 from both sides.

$$kx=-2x^{2}-5$$

The equation is in standard form.

$$xk=-2x^{2}-5$$

Divide both sides by x.

$$\frac{xk}{x}=\frac{-2x^{2}-5}{x}$$

Dividing by x undoes the multiplication by x.

$$k=\frac{-2x^{2}-5}{x}$$

Divide -2x^{2}-5 by x.

$$k=-2x-\frac{5}{x}$$

Solve for x (complex solution)

$$x=\frac{\sqrt{k^{2}-40}-k}{4}$$

$$x=\frac{-\sqrt{k^{2}-40}-k}{4}$$

Solve for x

$$x=\frac{\sqrt{k^{2}-40}-k}{4}$$

$$x=\frac{-\sqrt{k^{2}-40}-k}{4}\text{, }|k|\geq 2\sqrt{10}$$