What are the solutions to the equation 2x^2+4x-5=0 ?

The solutions to the equation 2x^2+4x-5=0 are x=(-2+sqrt(14))/2 ,x=-(2+sqrt(14))/2

Find the zeros of 2x^2+4x-5=0

The zeros of 2x^2+4x-5=0 are x=(-2+sqrt(14))/2 ,x=-(2+sqrt(14))/2

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2 x^{2} + 4 x - 5 = 0

x = \frac{- 2 + 14}{2}, x = - \frac{2 + 14}{2}

Decimal Notation

x = 0.87082 \dots, x = - 2.87082 \dots

Solution steps

2 x^{2} + 4 x - 5 = 0

Solve with the quadratic formula

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

4^{2} - 4 \cdot 2 (- 5) = 214

x_{1, 2} = \frac{- 4 \pm 2 14}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 4 + 2 14}{2 \cdot 2}, x_{2} = \frac{- 4 - 2 14}{2 \cdot 2}

x = \frac{- 4 + 2 14}{2 \cdot 2} : \frac{- 2 + 14}{2}

x = \frac{- 4 - 2 14}{2 \cdot 2} : - \frac{2 + 14}{2}

The solutions to the quadratic equation are:

x = \frac{- 2 + 14}{2}, x = - \frac{2 + 14}{2}

20 = b^{2} - b

8 x^{5} - 14 x^{4} - 22 x^{3} + 57 x^{2} - 35 x + 6

(- 1)^{2}

\frac{7}{12} + \frac{1}{3}

(- 216)^{\frac{2}{3}}

What are the solutions to the equation 2x^2+4x-5=0 ?
The solutions to the equation 2x^2+4x-5=0 are x=(-2+sqrt(14))/2 ,x=-(2+sqrt(14))/2
Find the zeros of 2x^2+4x-5=0
The zeros of 2x^2+4x-5=0 are x=(-2+sqrt(14))/2 ,x=-(2+sqrt(14))/2