Solve left(2x-8right)left(6-xright)=2

$$\left( 2x-8 \right) \left( 6-x \right) = 2$$

Solve for x

$$x=5$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 2x-8 by 6-x and combine like terms.

$$20x-2x^{2}-48=2$$

Subtract 2 from both sides.

$$20x-2x^{2}-48-2=0$$

Subtract 2 from -48 to get -50.

$$20x-2x^{2}-50=0$$

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.

$$-2x^{2}+20x-50=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 20 for b, and -50 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-20±\sqrt{20^{2}-4\left(-2\right)\left(-50\right)}}{2\left(-2\right)}$$

Square 20.

$$x=\frac{-20±\sqrt{400-4\left(-2\right)\left(-50\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-20±\sqrt{400+8\left(-50\right)}}{2\left(-2\right)}$$

Multiply 8 times -50.

$$x=\frac{-20±\sqrt{400-400}}{2\left(-2\right)}$$

Add 400 to -400.

$$x=\frac{-20±\sqrt{0}}{2\left(-2\right)}$$

Take the square root of 0.

$$x=-\frac{20}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=-\frac{20}{-4}$$

Divide -20 by -4.

$$x=5$$

Steps for Completing the Square

Use the distributive property to multiply 2x-8 by 6-x and combine like terms.

$$20x-2x^{2}-48=2$$

Add 48 to both sides.

$$20x-2x^{2}=2+48$$

Add 2 and 48 to get 50.

$$20x-2x^{2}=50$$

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.

$$-2x^{2}+20x=50$$

Divide both sides by -2.

$$\frac{-2x^{2}+20x}{-2}=\frac{50}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\frac{20}{-2}x=\frac{50}{-2}$$

Divide 20 by -2.

$$x^{2}-10x=\frac{50}{-2}$$

Divide 50 by -2.

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

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.

$$x=5$$