What are the solutions to the equation x+2=x^2-1 ?

The solutions to the equation x+2=x^2-1 are x=(1+sqrt(13))/2 ,x=(1-sqrt(13))/2

Find the zeros of x+2=x^2-1

The zeros of x+2=x^2-1 are x=(1+sqrt(13))/2 ,x=(1-sqrt(13))/2

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x + 2 = x^{2} - 1

x = \frac{1 + 13}{2}, x = \frac{1 - 13}{2}

Decimal Notation

x = 2.30277 \dots, x = - 1.30277 \dots

Solution steps

x + 2 = x^{2} - 1

Switch sides

x^{2} - 1 = x + 2

Move

2

to the left side

x^{2} - 3 = x

Move

x

to the left side

x^{2} - 3 - x = 0

Write in the standard form

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

x^{2} - x - 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot ( - 3 )}{2 \cdot 1}

(- 1)^{2} - 4 \cdot 1 \cdot (- 3) = 13

x_{1, 2} = \frac{- ( - 1 ) \pm 13}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 13}{2 \cdot 1}, x_{2} = \frac{- ( - 1 ) - 13}{2 \cdot 1}

x = \frac{- ( - 1 ) + 13}{2 \cdot 1} : \frac{1 + 13}{2}

x = \frac{- ( - 1 ) - 13}{2 \cdot 1} : \frac{1 - 13}{2}

The solutions to the quadratic equation are:

x = \frac{1 + 13}{2}, x = \frac{1 - 13}{2}

5 x + 3 = - 2 x^{2}

x, 2 x^{2} + 5 y^{2} = 0

16 y^{2} + 62 y - 45 = 0

y, y^{2} - t^{2} + 2 t y = C

24 x^{2} - 10 x - 3 = 0

What are the solutions to the equation x+2=x^2-1 ?
The solutions to the equation x+2=x^2-1 are x=(1+sqrt(13))/2 ,x=(1-sqrt(13))/2
Find the zeros of x+2=x^2-1
The zeros of x+2=x^2-1 are x=(1+sqrt(13))/2 ,x=(1-sqrt(13))/2