Solve -(H)^2/k+3=2

$$- \frac { ( H ) ^ { 2 } } { k + 3 } = 2$$

Solve for k

$$k=-\frac{H^{2}}{2}-3$$

$$H\neq 0$$

Steps for Solving Linear Equation

Variable k cannot be equal to -3 since division by zero is not defined. Multiply both sides of the equation by k+3.

$$-H^{2}=2\left(k+3\right)$$

Use the distributive property to multiply 2 by k+3.

$$-H^{2}=2k+6$$

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

$$2k+6=-H^{2}$$

Subtract 6 from both sides.

$$2k=-H^{2}-6$$

Divide both sides by 2.

$$\frac{2k}{2}=\frac{-H^{2}-6}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$k=\frac{-H^{2}-6}{2}$$

Divide -H^{2}-6 by 2.

$$k=-\frac{H^{2}}{2}-3$$

Variable k cannot be equal to -3.

$$k=-\frac{H^{2}}{2}-3\text{, }k\neq -3$$

Solve for H

$$H=\sqrt{-2k-6}$$

$$H=-\sqrt{-2k-6}\text{, }k<-3$$