Solve {l}{(x-sqrt{2})*(x+sqrt{2})=2x*(x-3)-2}{x^2-}

$$\left. \begin{array} { l } { ( x - \sqrt { 2 } ) \cdot ( x + \sqrt { 2 } ) = 2 x \cdot ( x - 3 ) - 2 } \\ { x ^ { 2 } - } \end{array} \right.$$

Solve for x

$$x=6$$

$$x=0$$

Steps Using the Quadratic Formula

Consider \left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

The square of \sqrt{2} is 2.

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

Use the distributive property to multiply 2x by x-3.

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

Subtract 2x^{2} from both sides.

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

Combine x^{2} and -2x^{2} to get -x^{2}.

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

Add 6x to both sides.

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

Add 2 to both sides.

$$-x^{2}-2+6x+2=0$$

Add -2 and 2 to get 0.

$$-x^{2}+6x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 6 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Take the square root of 6^{2}.

$$x=\frac{-6±6}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-6±6}{-2}$$

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

$$x=\frac{0}{-2}$$

Divide 0 by -2.

$$x=0$$

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

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

Divide -12 by -2.

$$x=6$$

The equation is now solved.

$$x=0 x=6$$

Steps for Completing the Square

Consider \left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

The square of \sqrt{2} is 2.

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

Use the distributive property to multiply 2x by x-3.

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

Subtract 2x^{2} from both sides.

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

Combine x^{2} and -2x^{2} to get -x^{2}.

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

Add 6x to both sides.

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

Add 2 to both sides.

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

Add -2 and 2 to get 0.

$$-x^{2}+6x=0$$

Divide both sides by -1.

$$\frac{-x^{2}+6x}{-1}=\frac{0}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{6}{-1}x=\frac{0}{-1}$$

Divide 6 by -1.

$$x^{2}-6x=\frac{0}{-1}$$

Divide 0 by -1.

$$x^{2}-6x=0$$

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.

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

Square -3.

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

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}.

$$\left(x-3\right)^{2}=9$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{9}$$

Simplify.

$$x-3=3 x-3=-3$$

Add 3 to both sides of the equation.

$$x=6 x=0$$