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

The solutions to the equation X^2=2+X are X=2,X=-1

Find the zeros of X^2=2+X

The zeros of X^2=2+X are X=2,X=-1

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X^{2} = 2 + X

X = 2, X = - 1

Solution steps

X^{2} = 2 + X

Move

X

to the left side

X^{2} - X = 2

Move

2

to the left side

X^{2} - X - 2 = 0

Solve with the quadratic formula

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

(- 1)^{2} - 4 \cdot 1 \cdot (- 2) = 3

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

X = 2, X = - 1

s (s + 2) = 0

2 a^{2} - 9 a - 5 = 0

x, x^{2} + y^{4} = 10

x^{2} - x = 132

x^{2} - 23 x + 102 = 0

What are the solutions to the equation X^2=2+X ?
The solutions to the equation X^2=2+X are X=2,X=-1
Find the zeros of X^2=2+X
The zeros of X^2=2+X are X=2,X=-1