What are the solutions to the equation 2x^2+4x+96=0 ?

The solutions to the equation 2x^2+4x+96=0 are x=-1+sqrt(47)i,x=-1-sqrt(47)i

Find the zeros of 2x^2+4x+96=0

The zeros of 2x^2+4x+96=0 are x=-1+sqrt(47)i,x=-1-sqrt(47)i

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2 x^{2} + 4 x + 96 = 0

x = - 1 + 47 i, x = - 1 - 47 i

Solution steps

2 x^{2} + 4 x + 96 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 2 \cdot 96}{2 \cdot 2}

Simplify

4^{2} - 4 \cdot 2 \cdot 96 : 447 i

x_{1, 2} = \frac{- 4 \pm 4 47 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 4 + 4 47 i}{2 \cdot 2}, x_{2} = \frac{- 4 - 4 47 i}{2 \cdot 2}

x = \frac{- 4 + 4 47 i}{2 \cdot 2} : - 1 + 47 i

x = \frac{- 4 - 4 47 i}{2 \cdot 2} : - 1 - 47 i

The solutions to the quadratic equation are:

x = - 1 + 47 i, x = - 1 - 47 i

x, 64 x^{2} - 48 x y + 9 y^{2} = 0

x^{2} (- 2) = 16

x^{2} + x + 32 = 0

y, (x + y)^{2} = x

r^{2} - 8 r - 22 = 6

What are the solutions to the equation 2x^2+4x+96=0 ?
The solutions to the equation 2x^2+4x+96=0 are x=-1+sqrt(47)i,x=-1-sqrt(47)i
Find the zeros of 2x^2+4x+96=0
The zeros of 2x^2+4x+96=0 are x=-1+sqrt(47)i,x=-1-sqrt(47)i