Solve y=2x^2+(k+3)x+(k+2)

$$y = 2 x ^ { 2 } + ( k + 3 ) x + ( k + 2 )$$

Solve for k

$$\left\{\begin{matrix}k=-\frac{2x^{2}+3x-y+2}{x+1}\text{, }&x\neq -1\\k\in \mathrm{R}\text{, }&y=1\text{ and }x=-1\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$y=2x^{2}+kx+3x+k+2$$

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

$$2x^{2}+kx+3x+k+2=y$$

Subtract 2x^{2} from both sides.

$$kx+3x+k+2=y-2x^{2}$$

Subtract 3x from both sides.

$$kx+k+2=y-2x^{2}-3x$$

Subtract 2 from both sides.

$$kx+k=y-2x^{2}-3x-2$$

Combine all terms containing k.

$$\left(x+1\right)k=y-2x^{2}-3x-2$$

The equation is in standard form.

$$\left(x+1\right)k=-2x^{2}-3x+y-2$$

Divide both sides by x+1.

$$\frac{\left(x+1\right)k}{x+1}=\frac{-2x^{2}-3x+y-2}{x+1}$$

Dividing by x+1 undoes the multiplication by x+1.

$$k=\frac{-2x^{2}-3x+y-2}{x+1}$$

Solve for x

$$x=\frac{\sqrt{8y+k^{2}-2k-7}-k-3}{4}$$

$$x=\frac{-\sqrt{8y+k^{2}-2k-7}-k-3}{4}\text{, }y\geq -\frac{k^{2}}{8}+\frac{k}{4}+\frac{7}{8}$$