Solve (x+1)^2/2-3x=K

$$\frac { ( x + 1 ) ^ { 2 } } { 2 - 3 x } = K$$

Solve for K

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

$$x\neq \frac{2}{3}$$

Steps for Solving Linear Equation

Multiply both sides of the equation by -3x+2.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

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

Use the distributive property to multiply K by -3x+2.

$$x^{2}+2x+1=-3Kx+2K$$

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

$$-3Kx+2K=x^{2}+2x+1$$

Combine all terms containing K.

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

The equation is in standard form.

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

Divide both sides by 2-3x.

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

Dividing by 2-3x undoes the multiplication by 2-3x.

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

Solve for x (complex solution)

$$x=\frac{\sqrt{K\left(9K+20\right)}-3K-2}{2}$$

$$x=\frac{-\sqrt{K\left(9K+20\right)}-3K-2}{2}$$

Solve for x

$$x=\frac{\sqrt{K\left(9K+20\right)}-3K-2}{2}$$

$$x=\frac{-\sqrt{K\left(9K+20\right)}-3K-2}{2}\text{, }K\leq -\frac{20}{9}\text{ or }K\geq 0$$