Solve 2x^2+4x-9=x^2h+2hx+k

$$2 x ^ { 2 } + 4 x - 9 = x ^ { 2 } h + 2 h x + k$$

Solve for h (complex solution)

$$\left\{\begin{matrix}h=-\frac{9+k-4x-2x^{2}}{x\left(x+2\right)}\text{, }&x\neq -2\text{ and }x\neq 0\\h\in \mathrm{C}\text{, }&\left(x=0\text{ or }x=-2\right)\text{ and }k=-9\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$x^{2}h+2hx+k=2x^{2}+4x-9$$

Subtract k from both sides.

$$x^{2}h+2hx=2x^{2}+4x-9-k$$

Combine all terms containing h.

$$\left(x^{2}+2x\right)h=2x^{2}+4x-9-k$$

The equation is in standard form.

$$\left(x^{2}+2x\right)h=2x^{2}+4x-k-9$$

Divide both sides by x^{2}+2x.

$$\frac{\left(x^{2}+2x\right)h}{x^{2}+2x}=\frac{2x^{2}+4x-k-9}{x^{2}+2x}$$

Dividing by x^{2}+2x undoes the multiplication by x^{2}+2x.

$$h=\frac{2x^{2}+4x-k-9}{x^{2}+2x}$$

Divide 2x^{2}+4x-9-k by x^{2}+2x.

$$h=\frac{2x^{2}+4x-k-9}{x\left(x+2\right)}$$

Solve for h

$$\left\{\begin{matrix}h=-\frac{9+k-4x-2x^{2}}{x\left(x+2\right)}\text{, }&x\neq -2\text{ and }x\neq 0\\h\in \mathrm{R}\text{, }&\left(x=0\text{ or }x=-2\right)\text{ and }k=-9\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$x^{2}h+2hx+k=2x^{2}+4x-9$$

Subtract k from both sides.

$$x^{2}h+2hx=2x^{2}+4x-9-k$$

Combine all terms containing h.

$$\left(x^{2}+2x\right)h=2x^{2}+4x-9-k$$

The equation is in standard form.

$$\left(x^{2}+2x\right)h=2x^{2}+4x-k-9$$

Divide both sides by x^{2}+2x.

$$\frac{\left(x^{2}+2x\right)h}{x^{2}+2x}=\frac{2x^{2}+4x-k-9}{x^{2}+2x}$$

Dividing by x^{2}+2x undoes the multiplication by x^{2}+2x.

$$h=\frac{2x^{2}+4x-k-9}{x^{2}+2x}$$

Divide 2x^{2}+4x-9-k by x^{2}+2x.

$$h=\frac{2x^{2}+4x-k-9}{x\left(x+2\right)}$$

Solve for k

$$k=-\left(hx^{2}-2x^{2}+2hx-4x+9\right)$$