What are the solutions to the equation 2x^2+4x+25=0 ?

The solutions to the equation 2x^2+4x+25=0 are x=-1+isqrt(23/2),x=-1-isqrt(23/2)

Find the zeros of 2x^2+4x+25=0

The zeros of 2x^2+4x+25=0 are x=-1+isqrt(23/2),x=-1-isqrt(23/2)

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2 x^{2} + 4 x + 25 = 0

x = - 1 + i \frac{23}{2}, x = - 1 - i \frac{23}{2}

Solution steps

2 x^{2} + 4 x + 25 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 2 \cdot 25}{2 \cdot 2}

Simplify

4^{2} - 4 \cdot 2 \cdot 25 : 246 i

x_{1, 2} = \frac{- 4 \pm 2 46 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 4 + 2 46 i}{2 \cdot 2}, x_{2} = \frac{- 4 - 2 46 i}{2 \cdot 2}

x = \frac{- 4 + 2 46 i}{2 \cdot 2} : - 1 + i \frac{23}{2}

x = \frac{- 4 - 2 46 i}{2 \cdot 2} : - 1 - i \frac{23}{2}

The solutions to the quadratic equation are:

x = - 1 + i \frac{23}{2}, x = - 1 - i \frac{23}{2}

(x - 4)^{2} = 49

2 (x + 9)^{2} = 40

x^{2} + 3 x + \frac{9}{4} = 0

4 a^{3} + 12 a^{2} - 64 a - 192

w + 9 < - 6 w + 3

What are the solutions to the equation 2x^2+4x+25=0 ?
The solutions to the equation 2x^2+4x+25=0 are x=-1+isqrt(23/2),x=-1-isqrt(23/2)
Find the zeros of 2x^2+4x+25=0
The zeros of 2x^2+4x+25=0 are x=-1+isqrt(23/2),x=-1-isqrt(23/2)