What are the solutions to the equation x^2+x=5 ?

The solutions to the equation x^2+x=5 are x=(-1+sqrt(21))/2 ,x=(-1-sqrt(21))/2

Find the zeros of x^2+x=5

The zeros of x^2+x=5 are x=(-1+sqrt(21))/2 ,x=(-1-sqrt(21))/2

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x^{2} + x = 5

x = \frac{- 1 + 21}{2}, x = \frac{- 1 - 21}{2}

Decimal Notation

x = 1.79128 \dots, x = - 2.79128 \dots

Solution steps

x^{2} + x = 5

Move

5

to the left side

x^{2} + x - 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 5 )}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot (- 5) = 21

x_{1, 2} = \frac{- 1 \pm 21}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 1 + 21}{2 \cdot 1}, x_{2} = \frac{- 1 - 21}{2 \cdot 1}

x = \frac{- 1 + 21}{2 \cdot 1} : \frac{- 1 + 21}{2}

x = \frac{- 1 - 21}{2 \cdot 1} : \frac{- 1 - 21}{2}

The solutions to the quadratic equation are:

x = \frac{- 1 + 21}{2}, x = \frac{- 1 - 21}{2}

(x + 2) (2 x + 1) - (x - 1)^{2} = 19

6 x^{2} + 2 x + 2 = 0

x^{2} + 8 x = - 20

x^{2} - 4 x - 12 = 0

5 x^{2} - 39 x - 4 = 4

What are the solutions to the equation x^2+x=5 ?
The solutions to the equation x^2+x=5 are x=(-1+sqrt(21))/2 ,x=(-1-sqrt(21))/2
Find the zeros of x^2+x=5
The zeros of x^2+x=5 are x=(-1+sqrt(21))/2 ,x=(-1-sqrt(21))/2