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

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

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

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

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

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

Decimal Notation

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

Solution steps

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

Switch sides

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}

24 = - 0.5 t^{2} + 3.2 t + 22

b^{2} = 34

50 = \frac{3 x ( 20 - x )}{2}

4 x^{2} + 36 x + 31

15 x^{2} - 6 = - 13 x

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