What are the solutions to the equation 0=x^2+2x-4 ?

The solutions to the equation 0=x^2+2x-4 are x=-1+sqrt(5),x=-1-sqrt(5)

Find the zeros of 0=x^2+2x-4

The zeros of 0=x^2+2x-4 are x=-1+sqrt(5),x=-1-sqrt(5)

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

x = - 1 + 5, x = - 1 - 5

Decimal Notation

x = 1.23606 \dots, x = - 3.23606 \dots

Solution steps

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

Switch sides

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

Solve with the quadratic formula

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

2^{2} - 4 \cdot 1 \cdot (- 4) = 25

x_{1, 2} = \frac{- 2 \pm 2 5}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 2 + 2 5}{2 \cdot 1}, x_{2} = \frac{- 2 - 2 5}{2 \cdot 1}

x = \frac{- 2 + 2 5}{2 \cdot 1} : - 1 + 5

x = \frac{- 2 - 2 5}{2 \cdot 1} : - 1 - 5

The solutions to the quadratic equation are:

x = - 1 + 5, x = - 1 - 5

3 x = - 1

14 - 4 y \leq 66

t^{2} - 12 t + 36

f = - 2 + 5 - 3

(\frac{1 + 5}{2})^{2} = 1

What are the solutions to the equation 0=x^2+2x-4 ?
The solutions to the equation 0=x^2+2x-4 are x=-1+sqrt(5),x=-1-sqrt(5)
Find the zeros of 0=x^2+2x-4
The zeros of 0=x^2+2x-4 are x=-1+sqrt(5),x=-1-sqrt(5)