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

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

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

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

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x^{2} = 2 x + 2

x = 1 + 3, x = 1 - 3

Decimal Notation

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

Solution steps

x^{2} = 2 x + 2

Move

2

to the left side

x^{2} - 2 = 2 x

Move

2 x

to the left side

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

Write in the standard form

a x^{2} + b x + c = 0

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

2 x^{2} - 8 x + 15 = 0

4 x^{2} - 30 = 70

4 x^{2} - 4 x + 1 = 0

5 n^{2} - 10 n + 5 = 0

t, 8 l e f t (x - 2) l e f t (x + \frac{2}{5})^{2} = 0

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