What are the solutions to the equation 2x^2+8x=-5 ?

The solutions to the equation 2x^2+8x=-5 are x=(-4+sqrt(6))/2 ,x=-(4+sqrt(6))/2

Find the zeros of 2x^2+8x=-5

The zeros of 2x^2+8x=-5 are x=(-4+sqrt(6))/2 ,x=-(4+sqrt(6))/2

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2 x^{2} + 8 x = - 5

x = \frac{- 4 + 6}{2}, x = - \frac{4 + 6}{2}

Decimal Notation

x = - 0.77525 \dots, x = - 3.22474 \dots

Solution steps

2 x^{2} + 8 x = - 5

Move

5

to the left side

2 x^{2} + 8 x + 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 2 \cdot 5}{2 \cdot 2}

8^{2} - 4 \cdot 2 \cdot 5 = 26

x_{1, 2} = \frac{- 8 \pm 2 6}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 8 + 2 6}{2 \cdot 2}, x_{2} = \frac{- 8 - 2 6}{2 \cdot 2}

x = \frac{- 8 + 2 6}{2 \cdot 2} : \frac{- 4 + 6}{2}

x = \frac{- 8 - 2 6}{2 \cdot 2} : - \frac{4 + 6}{2}

The solutions to the quadratic equation are:

x = \frac{- 4 + 6}{2}, x = - \frac{4 + 6}{2}

x = - x^{2} + 2

5 a^{2} - 10 a = 0

- 3 x^{2} + 6 x - 9 = 0

x, \frac{x ^{2}}{2} + \frac{z ^{2}}{3} = y

t, x^{2} p^{2} - 2 t x p + 2 x^{2} - t^{2} = 0

What are the solutions to the equation 2x^2+8x=-5 ?
The solutions to the equation 2x^2+8x=-5 are x=(-4+sqrt(6))/2 ,x=-(4+sqrt(6))/2
Find the zeros of 2x^2+8x=-5
The zeros of 2x^2+8x=-5 are x=(-4+sqrt(6))/2 ,x=-(4+sqrt(6))/2