What are the solutions to the equation 2x^2-2x-2=x^2 ?

The solutions to the equation 2x^2-2x-2=x^2 are x=1+sqrt(3),x=1-sqrt(3)

Find the zeros of 2x^2-2x-2=x^2

The zeros of 2x^2-2x-2=x^2 are x=1+sqrt(3),x=1-sqrt(3)

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2 x^{2} - 2 x - 2 = x^{2}

x = 1 + 3, x = 1 - 3

Decimal Notation

x = 2.73205 \dots, x = - 0.73205 \dots

Solution steps

2 x^{2} - 2 x - 2 = x^{2}

Move

x^{2}

to the left side

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

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot ( - 2 )}{2 \cdot 1}

(- 2)^{2} - 4 \cdot 1 \cdot (- 2) = 23

x_{1, 2} = \frac{- ( - 2 ) \pm 2 3}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 2 3}{2 \cdot 1}, x_{2} = \frac{- ( - 2 ) - 2 3}{2 \cdot 1}

x = \frac{- ( - 2 ) + 2 3}{2 \cdot 1} : 1 + 3

x = \frac{- ( - 2 ) - 2 3}{2 \cdot 1} : 1 - 3

The solutions to the quadratic equation are:

x = 1 + 3, x = 1 - 3

27 x^{2} + 19 x - 37 = 0

- 4 w^{2} + 16 w - 16 = 0

3 x^{2} - 36 x + 81 = 0

x^{2} - 12 = 7 x

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

What are the solutions to the equation 2x^2-2x-2=x^2 ?
The solutions to the equation 2x^2-2x-2=x^2 are x=1+sqrt(3),x=1-sqrt(3)
Find the zeros of 2x^2-2x-2=x^2
The zeros of 2x^2-2x-2=x^2 are x=1+sqrt(3),x=1-sqrt(3)