Solve 6-4x-x^2=x+4

$$6 - 4 x - x ^ { 2 } = x + 4$$

Solve for x

$$x=\frac{\sqrt{33}-5}{2}\approx 0.372281323$$

$$x=\frac{-\sqrt{33}-5}{2}\approx -5.372281323$$

Steps Using the Quadratic Formula

Subtract x from both sides.

$$6-4x-x^{2}-x=4$$

Combine -4x and -x to get -5x.

$$6-5x-x^{2}=4$$

Subtract 4 from both sides.

$$6-5x-x^{2}-4=0$$

Subtract 4 from 6 to get 2.

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

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.

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

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

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

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\left(-1\right)\times 2}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-5\right)±\sqrt{25+4\times 2}}{2\left(-1\right)}$$

Multiply 4 times 2.

$$x=\frac{-\left(-5\right)±\sqrt{25+8}}{2\left(-1\right)}$$

Add 25 to 8.

$$x=\frac{-\left(-5\right)±\sqrt{33}}{2\left(-1\right)}$$

The opposite of -5 is 5.

$$x=\frac{5±\sqrt{33}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{5±\sqrt{33}}{-2}$$

Now solve the equation x=\frac{5±\sqrt{33}}{-2} when ± is plus. Add 5 to \sqrt{33}.

$$x=\frac{\sqrt{33}+5}{-2}$$

Divide 5+\sqrt{33} by -2.

$$x=\frac{-\sqrt{33}-5}{2}$$

Now solve the equation x=\frac{5±\sqrt{33}}{-2} when ± is minus. Subtract \sqrt{33} from 5.

$$x=\frac{5-\sqrt{33}}{-2}$$

Divide 5-\sqrt{33} by -2.

$$x=\frac{\sqrt{33}-5}{2}$$

The equation is now solved.

$$x=\frac{-\sqrt{33}-5}{2} x=\frac{\sqrt{33}-5}{2}$$

Steps for Completing the Square

Subtract x from both sides.

$$6-4x-x^{2}-x=4$$

Combine -4x and -x to get -5x.

$$6-5x-x^{2}=4$$

Subtract 6 from both sides.

$$-5x-x^{2}=4-6$$

Subtract 6 from 4 to get -2.

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

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.

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

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide -5 by -1.

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

Divide -2 by -1.

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

Divide 5, the coefficient of the x term, by 2 to get \frac{5}{2}. Then add the square of \frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+5x+\frac{25}{4}=2+\frac{25}{4}$$

Add 2 to \frac{25}{4}.

$$x^{2}+5x+\frac{25}{4}=\frac{33}{4}$$

Factor x^{2}+5x+\frac{25}{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}.

$$\left(x+\frac{5}{2}\right)^{2}=\frac{33}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{2}\right)^{2}}=\sqrt{\frac{33}{4}}$$

Simplify.

$$x+\frac{5}{2}=\frac{\sqrt{33}}{2} x+\frac{5}{2}=-\frac{\sqrt{33}}{2}$$

Subtract \frac{5}{2} from both sides of the equation.

$$x=\frac{\sqrt{33}-5}{2} x=\frac{-\sqrt{33}-5}{2}$$