Solve (2)-x^2+8x+9=0

$$( 2 ) - x ^ { 2 } + 8 x + 9 = 0$$

Solve for x

$$x=3\sqrt{3}+4\approx 9.196152423$$

$$x=4-3\sqrt{3}\approx -1.196152423$$

Steps Using the Quadratic Formula

Add 2 and 9 to get 11.

$$11-x^{2}+8x=0$$

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.

$$-x^{2}+8x+11=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 8 for b, and 11 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\left(-1\right)\times 11}}{2\left(-1\right)}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\left(-1\right)\times 11}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-8±\sqrt{64+4\times 11}}{2\left(-1\right)}$$

Multiply 4 times 11.

$$x=\frac{-8±\sqrt{64+44}}{2\left(-1\right)}$$

Add 64 to 44.

$$x=\frac{-8±\sqrt{108}}{2\left(-1\right)}$$

Take the square root of 108.

$$x=\frac{-8±6\sqrt{3}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-8±6\sqrt{3}}{-2}$$

Now solve the equation x=\frac{-8±6\sqrt{3}}{-2} when ± is plus. Add -8 to 6\sqrt{3}.

$$x=\frac{6\sqrt{3}-8}{-2}$$

Divide -8+6\sqrt{3} by -2.

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

Now solve the equation x=\frac{-8±6\sqrt{3}}{-2} when ± is minus. Subtract 6\sqrt{3} from -8.

$$x=\frac{-6\sqrt{3}-8}{-2}$$

Divide -8-6\sqrt{3} by -2.

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

The equation is now solved.

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

Steps for Completing the Square

Add 2 and 9 to get 11.

$$11-x^{2}+8x=0$$

Subtract 11 from both sides. Anything subtracted from zero gives its negation.

$$-x^{2}+8x=-11$$

Divide both sides by -1.

$$\frac{-x^{2}+8x}{-1}=-\frac{11}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{8}{-1}x=-\frac{11}{-1}$$

Divide 8 by -1.

$$x^{2}-8x=-\frac{11}{-1}$$

Divide -11 by -1.

$$x^{2}-8x=11$$

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.

$$x^{2}-8x+\left(-4\right)^{2}=11+\left(-4\right)^{2}$$

Square -4.

$$x^{2}-8x+16=11+16$$

Add 11 to 16.

$$x^{2}-8x+16=27$$

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}.

$$\left(x-4\right)^{2}=27$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-4\right)^{2}}=\sqrt{27}$$

Simplify.

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

Add 4 to both sides of the equation.

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