Solve -4x-5=x^2

$$- 4 x - 5 = x ^ { 2 }$$

Solve for x (complex solution)

$$x=-2-i$$

$$x=-2+i$$

Steps Using the Quadratic Formula

Subtract x^{2} from both sides.

$$-4x-5-x^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-x^{2}-4x-5=0$$

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

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\left(-1\right)\left(-5\right)}}{2\left(-1\right)}$$

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\left(-1\right)\left(-5\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-4\right)±\sqrt{16+4\left(-5\right)}}{2\left(-1\right)}$$

Multiply 4 times -5.

$$x=\frac{-\left(-4\right)±\sqrt{16-20}}{2\left(-1\right)}$$

Add 16 to -20.

$$x=\frac{-\left(-4\right)±\sqrt{-4}}{2\left(-1\right)}$$

Take the square root of -4.

$$x=\frac{-\left(-4\right)±2i}{2\left(-1\right)}$$

The opposite of -4 is 4.

$$x=\frac{4±2i}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{4±2i}{-2}$$

Now solve the equation x=\frac{4±2i}{-2} when ± is plus. Add 4 to 2i.

$$x=\frac{4+2i}{-2}$$

Divide 4+2i by -2.

$$x=-2-i$$

Now solve the equation x=\frac{4±2i}{-2} when ± is minus. Subtract 2i from 4.

$$x=\frac{4-2i}{-2}$$

Divide 4-2i by -2.

$$x=-2+i$$

The equation is now solved.

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

Steps for Completing the Square

Subtract x^{2} from both sides.

$$-4x-5-x^{2}=0$$

Add 5 to both sides. Anything plus zero gives itself.

$$-4x-x^{2}=5$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-x^{2}-4x=5$$

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{4}{-1}\right)x=\frac{5}{-1}$$

Divide -4 by -1.

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

Divide 5 by -1.

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

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

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

Square 2.

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

Add -5 to 4.

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

Factor x^{2}+4x+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Subtract 2 from both sides of the equation.

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