Solve kx^2+(4k+2)x+4k+5=0

$$k x ^ { 2 } + ( 4 k + 2 ) x + 4 k + 5 = 0$$

Solve for k

$$k=-\frac{2x+5}{\left(x+2\right)^{2}}$$

$$x\neq -2$$

Steps for Solving Linear Equation

Use the distributive property to multiply 4k+2 by x.

$$kx^{2}+4kx+2x+4k+5=0$$

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

$$kx^{2}+4kx+4k+5=-2x$$

Subtract 5 from both sides.

$$kx^{2}+4kx+4k=-2x-5$$

Combine all terms containing k.

$$\left(x^{2}+4x+4\right)k=-2x-5$$

Divide both sides by x^{2}+4x+4.

$$\frac{\left(x^{2}+4x+4\right)k}{x^{2}+4x+4}=\frac{-2x-5}{x^{2}+4x+4}$$

Dividing by x^{2}+4x+4 undoes the multiplication by x^{2}+4x+4.

$$k=\frac{-2x-5}{x^{2}+4x+4}$$

Divide -2x-5 by x^{2}+4x+4.

$$k=-\frac{2x+5}{\left(x+2\right)^{2}}$$

Solve for x (complex solution)

$$\left\{\begin{matrix}x=\frac{-2k+\sqrt{1-k}-1}{k}\text{; }x=-\frac{2k+\sqrt{1-k}+1}{k}\text{, }&k\neq 0\\x=-\frac{5}{2}\text{, }&k=0\end{matrix}\right.$$

Solve for x

$$\left\{\begin{matrix}x=\frac{-2k+\sqrt{1-k}-1}{k}\text{; }x=-\frac{2k+\sqrt{1-k}+1}{k}\text{, }&k\neq 0\text{ and }k\leq 1\\x=-\frac{5}{2}\text{, }&k=0\end{matrix}\right.$$