Solve 1/4x^2+x+2-(-1)/x-(-2)=-1

$$\frac { \frac { 1 } { 4 } x ^ { 2 } + x + 2 - ( - 1 ) } { x - ( - 2 ) } = - 1$$

Solve for x (complex solution)

$$x=-4+2i$$

$$x=-4-2i$$

Steps Using the Quadratic Formula

Variable x cannot be equal to -2 since division by zero is not defined. Multiply both sides of the equation by x+2.

$$\frac{1}{4}x^{2}+x+2-\left(-1\right)=-\left(x+2\right)$$

The opposite of -1 is 1.

$$\frac{1}{4}x^{2}+x+2+1=-\left(x+2\right)$$

To find the opposite of x+2, find the opposite of each term.

$$\frac{1}{4}x^{2}+x+2+1=-x-2$$

Add 2 and 1 to get 3.

$$\frac{1}{4}x^{2}+x+3=-x-2$$

Add x to both sides.

$$\frac{1}{4}x^{2}+x+3+x=-2$$

Combine x and x to get 2x.

$$\frac{1}{4}x^{2}+2x+3=-2$$

Add 2 to both sides.

$$\frac{1}{4}x^{2}+2x+3+2=0$$

Add 3 and 2 to get 5.

$$\frac{1}{4}x^{2}+2x+5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{4} for a, 2 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}-4\times \frac{1}{4}\times 5}}{2\times \frac{1}{4}}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\times \frac{1}{4}\times 5}}{2\times \frac{1}{4}}$$

Multiply -4 times \frac{1}{4}.

$$x=\frac{-2±\sqrt{4-5}}{2\times \frac{1}{4}}$$

Add 4 to -5.

$$x=\frac{-2±\sqrt{-1}}{2\times \frac{1}{4}}$$

Take the square root of -1.

$$x=\frac{-2±i}{2\times \frac{1}{4}}$$

Multiply 2 times \frac{1}{4}.

$$x=\frac{-2±i}{\frac{1}{2}}$$

Now solve the equation x=\frac{-2±i}{\frac{1}{2}} when ± is plus. Add -2 to i.

$$x=\frac{-2+i}{\frac{1}{2}}$$

Divide -2+i by \frac{1}{2} by multiplying -2+i by the reciprocal of \frac{1}{2}.

$$x=-4+2i$$

Now solve the equation x=\frac{-2±i}{\frac{1}{2}} when ± is minus. Subtract i from -2.

$$x=\frac{-2-i}{\frac{1}{2}}$$

Divide -2-i by \frac{1}{2} by multiplying -2-i by the reciprocal of \frac{1}{2}.

$$x=-4-2i$$

The equation is now solved.

$$x=-4+2i x=-4-2i$$

Steps for Completing the Square

Variable x cannot be equal to -2 since division by zero is not defined. Multiply both sides of the equation by x+2.

$$\frac{1}{4}x^{2}+x+2-\left(-1\right)=-\left(x+2\right)$$

The opposite of -1 is 1.

$$\frac{1}{4}x^{2}+x+2+1=-\left(x+2\right)$$

To find the opposite of x+2, find the opposite of each term.

$$\frac{1}{4}x^{2}+x+2+1=-x-2$$

Add 2 and 1 to get 3.

$$\frac{1}{4}x^{2}+x+3=-x-2$$

Add x to both sides.

$$\frac{1}{4}x^{2}+x+3+x=-2$$

Combine x and x to get 2x.

$$\frac{1}{4}x^{2}+2x+3=-2$$

Subtract 3 from both sides.

$$\frac{1}{4}x^{2}+2x=-2-3$$

Subtract 3 from -2 to get -5.

$$\frac{1}{4}x^{2}+2x=-5$$

Multiply both sides by 4.

$$\frac{\frac{1}{4}x^{2}+2x}{\frac{1}{4}}=-\frac{5}{\frac{1}{4}}$$

Dividing by \frac{1}{4} undoes the multiplication by \frac{1}{4}.

$$x^{2}+\frac{2}{\frac{1}{4}}x=-\frac{5}{\frac{1}{4}}$$

Divide 2 by \frac{1}{4} by multiplying 2 by the reciprocal of \frac{1}{4}.

$$x^{2}+8x=-\frac{5}{\frac{1}{4}}$$

Divide -5 by \frac{1}{4} by multiplying -5 by the reciprocal of \frac{1}{4}.

$$x^{2}+8x=-20$$

Divide 8, the coefficient of the x term, by 2 to get 4. Then add the square of 4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+8x+4^{2}=-20+4^{2}$$

Square 4.

$$x^{2}+8x+16=-20+16$$

Add -20 to 16.

$$x^{2}+8x+16=-4$$

Factor x^{2}+8x+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x+4=2i x+4=-2i$$

Subtract 4 from both sides of the equation.

$$x=-4+2i x=-4-2i$$