Solve x^2+2(m-1)x+(m+5)=0quad[2012]

$$x ^ { 2 } + 2 ( m - 1 ) x + ( m + 5 ) = 0 \quad [ 2012 ]$$

Solve for m

$$m=-\frac{x^{2}-2x+5}{2x+1}$$

$$x\neq -\frac{1}{2}$$

Steps for Solving Linear Equation

Use the distributive property to multiply 2 by m-1.

$$x^{2}+\left(2m-2\right)x+m+5=0\times 2012$$

Use the distributive property to multiply 2m-2 by x.

$$x^{2}+2mx-2x+m+5=0\times 2012$$

Multiply 0 and 2012 to get 0.

$$x^{2}+2mx-2x+m+5=0$$

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

$$2mx-2x+m+5=-x^{2}$$

Add 2x to both sides.

$$2mx+m+5=-x^{2}+2x$$

Subtract 5 from both sides.

$$2mx+m=-x^{2}+2x-5$$

Combine all terms containing m.

$$\left(2x+1\right)m=-x^{2}+2x-5$$

Divide both sides by 2x+1.

$$\frac{\left(2x+1\right)m}{2x+1}=\frac{-x^{2}+2x-5}{2x+1}$$

Dividing by 2x+1 undoes the multiplication by 2x+1.

$$m=\frac{-x^{2}+2x-5}{2x+1}$$

Solve for x (complex solution)

$$x=\sqrt{\left(m-4\right)\left(m+1\right)}-m+1$$

$$x=-\sqrt{\left(m-4\right)\left(m+1\right)}-m+1$$

Solve for x

$$x=\sqrt{\left(m-4\right)\left(m+1\right)}-m+1$$

$$x=-\sqrt{\left(m-4\right)\left(m+1\right)}-m+1\text{, }m\leq -1\text{ or }m\geq 4$$