What are the solutions to the equation x^2+x=3 ?

The solutions to the equation x^2+x=3 are x=(-1+sqrt(13))/2 ,x=(-1-sqrt(13))/2

Find the zeros of x^2+x=3

The zeros of x^2+x=3 are x=(-1+sqrt(13))/2 ,x=(-1-sqrt(13))/2

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x^{2} + x = 3

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

Decimal Notation

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

Solution steps

x^{2} + x = 3

Move

3

to the left side

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}

x, x^{2} - 3 k x + 3 k - 7 = 0

(x + 5) (x + 2) = 0

3 x^{2} + 7 = 301

x^{2} + 7 x = 68.75

2 x^{2} + 16 x - 5

What are the solutions to the equation x^2+x=3 ?
The solutions to the equation x^2+x=3 are x=(-1+sqrt(13))/2 ,x=(-1-sqrt(13))/2
Find the zeros of x^2+x=3
The zeros of x^2+x=3 are x=(-1+sqrt(13))/2 ,x=(-1-sqrt(13))/2