What are the solutions to the equation 2x^2+4x+18=0 ?

The solutions to the equation 2x^2+4x+18=0 are x=-1+2sqrt(2)i,x=-1-2sqrt(2)i

Find the zeros of 2x^2+4x+18=0

The zeros of 2x^2+4x+18=0 are x=-1+2sqrt(2)i,x=-1-2sqrt(2)i

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2 x^{2} + 4 x + 18 = 0

x = - 1 + 22 i, x = - 1 - 22 i

Solution steps

2 x^{2} + 4 x + 18 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 2 \cdot 18}{2 \cdot 2}

Simplify

4^{2} - 4 \cdot 2 \cdot 18 : 82 i

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

Separate the solutions

x_{1} = \frac{- 4 + 8 2 i}{2 \cdot 2}, x_{2} = \frac{- 4 - 8 2 i}{2 \cdot 2}

x = \frac{- 4 + 8 2 i}{2 \cdot 2} : - 1 + 22 i

x = \frac{- 4 - 8 2 i}{2 \cdot 2} : - 1 - 22 i

The solutions to the quadratic equation are:

x = - 1 + 22 i, x = - 1 - 22 i

5^{2} + b^{2} = 1 2^{2}

x^{2} + 3 x - 6 = 8

d^{2} + d - 72 = 0

x^{2} - x + 56 = 0

6 x^{2} - 19 x + 8 = 0

What are the solutions to the equation 2x^2+4x+18=0 ?
The solutions to the equation 2x^2+4x+18=0 are x=-1+2sqrt(2)i,x=-1-2sqrt(2)i
Find the zeros of 2x^2+4x+18=0
The zeros of 2x^2+4x+18=0 are x=-1+2sqrt(2)i,x=-1-2sqrt(2)i