Question

HELP SOLVE THANK YOU!

Answer 1

Let's first move everything, the constants and variables to one side

    $x^2-2=2x$ ⇒ $x^2-2x-2=0$

The second form is in the general form of the quadratic equation:

 ⇒ $ax^2+bx+c = 0$

• a ⇒ 1
• b ⇒ -2
• c ⇒ -2

To find the value of x, we can use the quadratic formula:

  ⇒ look at the diagram attached

  $x = \frac{-b-\sqrt{b^2-4ac} }{2a} =\frac{2-\sqrt{(-2)^2-4(1)(-2)} }{2*1} =\frac{2-\sqrt{4+8} }{2} =\frac{2-\sqrt{12} }{2} =\frac{2-2\sqrt{3}}{2} =1-\sqrt{3}$

  $x = \frac{-b+\sqrt{b^2-4ac} }{2a} =\frac{2+\sqrt{(-2)^2-4(1)(-2)} }{2*1} =\frac{2+\sqrt{4+8} }{2} =\frac{2+\sqrt{12} }{2} =\frac{2+2\sqrt{3}}{2} =1+\sqrt{3}$  

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                  Answer

To answer the question, just answer in the order I solved it, because the question asks you to put the answer in ascending order which means from least to greatest.

Hope that helps!

Answer 2

Answer:

Solution 1: 1+1$\sqrt{3}$

Solution 2: 1-$1\sqrt{3}$

Step-by-step explanation:

x^2-2=2x

x^2-2x-2=0

Use quadratic equation.

(-b+-$\sqrt{b^2-4ac}$)/2a

(2+-$\sqrt{(-2)^2-4*1*-2}$)/2

(2+-$\sqrt{(4+8})$)/2

(2+-$\sqrt{12}$)/2

Solution 1: (2+2$\sqrt{3}$)/2=1+1$\sqrt{3}$

Solution 2: 1-$1\sqrt{3}$