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

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

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

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

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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

2 x^{2} + 4 x = 5

Move

5

to the left side

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}

- \frac{n}{2} + 8 > - 4

\frac{1}{3} (x + 3) = \frac{2}{5} (x + 2)

- 317 - 417

\frac{\frac{1}{2}}{\frac{3}{2}}

6 + 9

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