What are the solutions to the equation x^2+2x-5=0 ?

The solutions to the equation x^2+2x-5=0 are x=-1+sqrt(6),x=-1-sqrt(6)

Find the zeros of x^2+2x-5=0

The zeros of x^2+2x-5=0 are x=-1+sqrt(6),x=-1-sqrt(6)

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x^{2} + 2 x - 5 = 0

x = - 1 + 6, x = - 1 - 6

Decimal Notation

x = 1.44948 \dots, x = - 3.44948 \dots

Solution steps

x^{2} + 2 x - 5 = 0

Solve with the quadratic formula

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

2^{2} - 4 \cdot 1 \cdot (- 5) = 26

x_{1, 2} = \frac{- 2 \pm 2 6}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 2 + 2 6}{2 \cdot 1}, x_{2} = \frac{- 2 - 2 6}{2 \cdot 1}

x = \frac{- 2 + 2 6}{2 \cdot 1} : - 1 + 6

x = \frac{- 2 - 2 6}{2 \cdot 1} : - 1 - 6

The solutions to the quadratic equation are:

x = - 1 + 6, x = - 1 - 6

\frac{x}{7} - \frac{x}{2} = 5

- 5 - 1

- 4 (x + 3) + 57 = - 35

x^{2} - 17 x + \frac{289}{4}

a = 70 - 14.99 (5)

What are the solutions to the equation x^2+2x-5=0 ?
The solutions to the equation x^2+2x-5=0 are x=-1+sqrt(6),x=-1-sqrt(6)
Find the zeros of x^2+2x-5=0
The zeros of x^2+2x-5=0 are x=-1+sqrt(6),x=-1-sqrt(6)