What are the solutions to the equation r^2+18r+82=0 ?

The solutions to the equation r^2+18r+82=0 are r=-9+i,r=-9-i

Find the zeros of r^2+18r+82=0

The zeros of r^2+18r+82=0 are r=-9+i,r=-9-i

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

r^{2} + 18 r + 82 = 0

r = - 9 + i, r = - 9 - i

Solution steps

r^{2} + 18 r + 82 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 18 \pm 1 8 ^{2} - 4 \cdot 1 \cdot 82}{2 \cdot 1}

Simplify

1 8^{2} - 4 \cdot 1 \cdot 82 : 2 i

r_{1, 2} = \frac{- 18 \pm 2 i}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- 18 + 2 i}{2 \cdot 1}, r_{2} = \frac{- 18 - 2 i}{2 \cdot 1}

r = \frac{- 18 + 2 i}{2 \cdot 1} : - 9 + i

r = \frac{- 18 - 2 i}{2 \cdot 1} : - 9 - i

The solutions to the quadratic equation are:

r = - 9 + i, r = - 9 - i

\frac{4}{5} \cdot 4

(2 x - 15) (5 x + 16) = 0

\frac{w}{6} + \frac{w}{8} = \frac{7}{6}

36 a^{2} - 49 b^{2}

- 3 x + 1 > 13

What are the solutions to the equation r^2+18r+82=0 ?
The solutions to the equation r^2+18r+82=0 are r=-9+i,r=-9-i
Find the zeros of r^2+18r+82=0
The zeros of r^2+18r+82=0 are r=-9+i,r=-9-i