Solve (x-2)(x-8)=-9

$$( x - 2 ) ( x - 8 ) = - 9$$

Solve for x

$$x=5$$

Steps Using the Quadratic Formula

Use the distributive property to multiply x-2 by x-8 and combine like terms.

$$x^{2}-10x+16=-9$$

Add 9 to both sides.

$$x^{2}-10x+16+9=0$$

Add 16 and 9 to get 25.

$$x^{2}-10x+25=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -10 for b, and 25 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 25}}{2}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times 25}}{2}$$

Multiply -4 times 25.

$$x=\frac{-\left(-10\right)±\sqrt{100-100}}{2}$$

Add 100 to -100.

$$x=\frac{-\left(-10\right)±\sqrt{0}}{2}$$

Take the square root of 0.

$$x=-\frac{-10}{2}$$

The opposite of -10 is 10.

$$x=\frac{10}{2}$$

Divide 10 by 2.

$$x=5$$

Steps for Completing the Square

Use the distributive property to multiply x-2 by x-8 and combine like terms.

$$x^{2}-10x+16=-9$$

Subtract 16 from both sides.

$$x^{2}-10x=-9-16$$

Subtract 16 from -9 to get -25.

$$x^{2}-10x=-25$$

Divide -10, the coefficient of the x term, by 2 to get -5. Then add the square of -5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-10x+\left(-5\right)^{2}=-25+\left(-5\right)^{2}$$

Square -5.

$$x^{2}-10x+25=-25+25$$

Add -25 to 25.

$$x^{2}-10x+25=0$$

Factor x^{2}-10x+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-5\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{0}$$

Simplify.

$$x-5=0 x-5=0$$

Add 5 to both sides of the equation.

$$x=5 x=5$$

The equation is now solved. Solutions are the same.

$$x=5$$