Solve (3k+1)x^2+4kx+5=0

$$( 3 k + 1 ) x ^ { 2 } + 4 k x + 5 = 0$$

Solve for k

$$k=-\frac{x^{2}+5}{x\left(3x+4\right)}$$

$$x\neq -\frac{4}{3}\text{ and }x\neq 0$$

Steps for Solving Linear Equation

Use the distributive property to multiply 3k+1 by x^{2}.

$$3kx^{2}+x^{2}+4kx+5=0$$

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

$$3kx^{2}+4kx+5=-x^{2}$$

Subtract 5 from both sides.

$$3kx^{2}+4kx=-x^{2}-5$$

Combine all terms containing k.

$$\left(3x^{2}+4x\right)k=-x^{2}-5$$

Divide both sides by 3x^{2}+4x.

$$\frac{\left(3x^{2}+4x\right)k}{3x^{2}+4x}=\frac{-x^{2}-5}{3x^{2}+4x}$$

Dividing by 3x^{2}+4x undoes the multiplication by 3x^{2}+4x.

$$k=\frac{-x^{2}-5}{3x^{2}+4x}$$

Divide -x^{2}-5 by 3x^{2}+4x.

$$k=-\frac{x^{2}+5}{x\left(3x+4\right)}$$

Solve for x (complex solution)

$$\left\{\begin{matrix}x=\frac{\sqrt{4k^{2}-15k-5}-2k}{3k+1}\text{; }x=-\frac{\sqrt{4k^{2}-15k-5}+2k}{3k+1}\text{, }&k\neq -\frac{1}{3}\\x=\frac{15}{4}\text{, }&k=-\frac{1}{3}\end{matrix}\right.$$

Solve for x

$$\left\{\begin{matrix}x=\frac{\sqrt{4k^{2}-15k-5}-2k}{3k+1}\text{; }x=-\frac{\sqrt{4k^{2}-15k-5}+2k}{3k+1}\text{, }&\left(k\neq -\frac{1}{3}\text{ and }k\leq \frac{15-\sqrt{305}}{8}\right)\text{ or }k\geq \frac{\sqrt{305}+15}{8}\\x=\frac{15}{4}\text{, }&k=-\frac{1}{3}\end{matrix}\right.$$