Solve 1/2(x-6)(2x-8)=1

$$\frac{ 1 }{ 2 } (x-6)(2x-8)=1$$

Solve for x

$$x=\sqrt{2}+5\approx 6.414213562$$

$$x=5-\sqrt{2}\approx 3.585786438$$

Steps Using the Quadratic Formula

Use the distributive property to multiply \frac{1}{2} by x-6.

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

Use the distributive property to multiply \frac{1}{2}x-3 by 2x-8 and combine like terms.

$$x^{2}-10x+24=1$$

Subtract 1 from both sides.

$$x^{2}-10x+24-1=0$$

Subtract 1 from 24 to get 23.

$$x^{2}-10x+23=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -10 for b, and 23 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 23}}{2}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times 23}}{2}$$

Multiply -4 times 23.

$$x=\frac{-\left(-10\right)±\sqrt{100-92}}{2}$$

Add 100 to -92.

$$x=\frac{-\left(-10\right)±\sqrt{8}}{2}$$

Take the square root of 8.

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

The opposite of -10 is 10.

$$x=\frac{10±2\sqrt{2}}{2}$$

Now solve the equation x=\frac{10±2\sqrt{2}}{2} when ± is plus. Add 10 to 2\sqrt{2}.

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

Divide 10+2\sqrt{2} by 2.

$$x=\sqrt{2}+5$$

Now solve the equation x=\frac{10±2\sqrt{2}}{2} when ± is minus. Subtract 2\sqrt{2} from 10.

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

Divide 10-2\sqrt{2} by 2.

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

The equation is now solved.

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

Steps for Completing the Square

Use the distributive property to multiply \frac{1}{2} by x-6.

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

Use the distributive property to multiply \frac{1}{2}x-3 by 2x-8 and combine like terms.

$$x^{2}-10x+24=1$$

Subtract 24 from both sides.

$$x^{2}-10x=1-24$$

Subtract 24 from 1 to get -23.

$$x^{2}-10x=-23$$

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.

$$x^{2}-10x+\left(-5\right)^{2}=-23+\left(-5\right)^{2}$$

Square -5.

$$x^{2}-10x+25=-23+25$$

Add -23 to 25.

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add 5 to both sides of the equation.

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