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

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

Solve for x

$$x=2\sqrt{2}+3\approx 5.828427125$$

$$x=3-2\sqrt{2}\approx 0.171572875$$

Steps Using the Quadratic Formula

Multiply both sides by 2.

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

Use the distributive property to multiply 2x by 6-x.

$$12x-2x^{2}=2$$

Subtract 2 from both sides.

$$12x-2x^{2}-2=0$$

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.

$$-2x^{2}+12x-2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 12 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-12±\sqrt{12^{2}-4\left(-2\right)\left(-2\right)}}{2\left(-2\right)}$$

Square 12.

$$x=\frac{-12±\sqrt{144-4\left(-2\right)\left(-2\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-12±\sqrt{144+8\left(-2\right)}}{2\left(-2\right)}$$

Multiply 8 times -2.

$$x=\frac{-12±\sqrt{144-16}}{2\left(-2\right)}$$

Add 144 to -16.

$$x=\frac{-12±\sqrt{128}}{2\left(-2\right)}$$

Take the square root of 128.

$$x=\frac{-12±8\sqrt{2}}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-12±8\sqrt{2}}{-4}$$

Now solve the equation x=\frac{-12±8\sqrt{2}}{-4} when ± is plus. Add -12 to 8\sqrt{2}.

$$x=\frac{8\sqrt{2}-12}{-4}$$

Divide -12+8\sqrt{2} by -4.

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

Now solve the equation x=\frac{-12±8\sqrt{2}}{-4} when ± is minus. Subtract 8\sqrt{2} from -12.

$$x=\frac{-8\sqrt{2}-12}{-4}$$

Divide -12-8\sqrt{2} by -4.

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

The equation is now solved.

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

Steps for Completing the Square

Multiply both sides by 2.

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

Use the distributive property to multiply 2x by 6-x.

$$12x-2x^{2}=2$$

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.

$$-2x^{2}+12x=2$$

Divide both sides by -2.

$$\frac{-2x^{2}+12x}{-2}=\frac{2}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\frac{12}{-2}x=\frac{2}{-2}$$

Divide 12 by -2.

$$x^{2}-6x=\frac{2}{-2}$$

Divide 2 by -2.

$$x^{2}-6x=-1$$

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-6x+\left(-3\right)^{2}=-1+\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=-1+9$$

Add -1 to 9.

$$x^{2}-6x+9=8$$

Factor x^{2}-6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{8}$$

Simplify.

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

Add 3 to both sides of the equation.

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