What are the solutions to the equation 2x^2+16x+33=0 ?

The solutions to the equation 2x^2+16x+33=0 are x=-4+i(sqrt(2))/2 ,x=-4-i(sqrt(2))/2

Find the zeros of 2x^2+16x+33=0

The zeros of 2x^2+16x+33=0 are x=-4+i(sqrt(2))/2 ,x=-4-i(sqrt(2))/2

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2 x^{2} + 16 x + 33 = 0

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

Solution steps

2 x^{2} + 16 x + 33 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 16 \pm 1 6 ^{2} - 4 \cdot 2 \cdot 33}{2 \cdot 2}

Simplify

1 6^{2} - 4 \cdot 2 \cdot 33 : 22 i

x_{1, 2} = \frac{- 16 \pm 2 2 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 16 + 2 2 i}{2 \cdot 2}, x_{2} = \frac{- 16 - 2 2 i}{2 \cdot 2}

x = \frac{- 16 + 2 2 i}{2 \cdot 2} : - 4 + i \frac{2}{2}

x = \frac{- 16 - 2 2 i}{2 \cdot 2} : - 4 - i \frac{2}{2}

The solutions to the quadratic equation are:

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

\frac{x ^{6}}{x ^{3}}

64 - y \geq - 128

504 m^{2} - 840 m + 294

\frac{2 π}{\frac{7}{4}}

- 11 \leq \frac{1}{2} z + 1 \leq 6

What are the solutions to the equation 2x^2+16x+33=0 ?
The solutions to the equation 2x^2+16x+33=0 are x=-4+i(sqrt(2))/2 ,x=-4-i(sqrt(2))/2
Find the zeros of 2x^2+16x+33=0
The zeros of 2x^2+16x+33=0 are x=-4+i(sqrt(2))/2 ,x=-4-i(sqrt(2))/2