What are the solutions to the equation x^2+2x-0.5=0 ?

The solutions to the equation x^2+2x-0.5=0 are x=(-2+sqrt(6))/2 ,x=-(2+sqrt(6))/2

Find the zeros of x^2+2x-0.5=0

The zeros of x^2+2x-0.5=0 are x=(-2+sqrt(6))/2 ,x=-(2+sqrt(6))/2

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x^{2} + 2 x - 0.5 = 0

x = \frac{- 2 + 6}{2}, x = - \frac{2 + 6}{2}

Decimal Notation

x = 0.22474 \dots, x = - 2.22474 \dots

Solution steps

x^{2} + 2 x - 0.5 = 0

Multiply both sides by

10

10 x^{2} + 20 x - 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 \cdot 10 ( - 5 )}{2 \cdot 10}

2 0^{2} - 4 \cdot 10 (- 5) = 106

x_{1, 2} = \frac{- 20 \pm 10 6}{2 \cdot 10}

Separate the solutions

x_{1} = \frac{- 20 + 10 6}{2 \cdot 10}, x_{2} = \frac{- 20 - 10 6}{2 \cdot 10}

x = \frac{- 20 + 10 6}{2 \cdot 10} : \frac{- 2 + 6}{2}

x = \frac{- 20 - 10 6}{2 \cdot 10} : - \frac{2 + 6}{2}

The solutions to the quadratic equation are:

x = \frac{- 2 + 6}{2}, x = - \frac{2 + 6}{2}

(x + 2)^{2} + x^{2} = 74

(3 y + 17)^{2} = 50

w^{2} = 8 (w - 4)

a, a b^{2} + a^{2} b + 3 = ab

4 x^{2} + 20 = 4

What are the solutions to the equation x^2+2x-0.5=0 ?
The solutions to the equation x^2+2x-0.5=0 are x=(-2+sqrt(6))/2 ,x=-(2+sqrt(6))/2
Find the zeros of x^2+2x-0.5=0
The zeros of x^2+2x-0.5=0 are x=(-2+sqrt(6))/2 ,x=-(2+sqrt(6))/2