Solve x=x^2-7x-2

$$x = x ^ { 2 } - 7 x - 2$$

Solve for x

$$x=3\sqrt{2}+4\approx 8.242640687$$

$$x=4-3\sqrt{2}\approx -0.242640687$$

Steps Using the Quadratic Formula

Subtract x^{2} from both sides.

$$x-x^{2}=-7x-2$$

Add 7x to both sides.

$$x-x^{2}+7x=-2$$

Combine x and 7x to get 8x.

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

Add 2 to both sides.

$$8x-x^{2}+2=0$$

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.

$$-x^{2}+8x+2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 8 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\left(-1\right)\times 2}}{2\left(-1\right)}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\left(-1\right)\times 2}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-8±\sqrt{64+4\times 2}}{2\left(-1\right)}$$

Multiply 4 times 2.

$$x=\frac{-8±\sqrt{64+8}}{2\left(-1\right)}$$

Add 64 to 8.

$$x=\frac{-8±\sqrt{72}}{2\left(-1\right)}$$

Take the square root of 72.

$$x=\frac{-8±6\sqrt{2}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-8±6\sqrt{2}}{-2}$$

Now solve the equation x=\frac{-8±6\sqrt{2}}{-2} when ± is plus. Add -8 to 6\sqrt{2}.

$$x=\frac{6\sqrt{2}-8}{-2}$$

Divide -8+6\sqrt{2} by -2.

$$x=4-3\sqrt{2}$$

Now solve the equation x=\frac{-8±6\sqrt{2}}{-2} when ± is minus. Subtract 6\sqrt{2} from -8.

$$x=\frac{-6\sqrt{2}-8}{-2}$$

Divide -8-6\sqrt{2} by -2.

$$x=3\sqrt{2}+4$$

The equation is now solved.

$$x=4-3\sqrt{2} x=3\sqrt{2}+4$$

Steps for Completing the Square

Subtract x^{2} from both sides.

$$x-x^{2}=-7x-2$$

Add 7x to both sides.

$$x-x^{2}+7x=-2$$

Combine x and 7x to get 8x.

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

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.

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

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide 8 by -1.

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

Divide -2 by -1.

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

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.

$$x^{2}-8x+\left(-4\right)^{2}=2+\left(-4\right)^{2}$$

Square -4.

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

Add 2 to 16.

$$x^{2}-8x+16=18$$

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}.

$$\left(x-4\right)^{2}=18$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-4\right)^{2}}=\sqrt{18}$$

Simplify.

$$x-4=3\sqrt{2} x-4=-3\sqrt{2}$$

Add 4 to both sides of the equation.

$$x=3\sqrt{2}+4 x=4-3\sqrt{2}$$