What are the solutions to the equation r^2+8r+32=0 ?

The solutions to the equation r^2+8r+32=0 are r=-4+4i,r=-4-4i

Find the zeros of r^2+8r+32=0

The zeros of r^2+8r+32=0 are r=-4+4i,r=-4-4i

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

r^{2} + 8 r + 32 = 0

r = - 4 + 4 i, r = - 4 - 4 i

Solution steps

r^{2} + 8 r + 32 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 1 \cdot 32}{2 \cdot 1}

Simplify

8^{2} - 4 \cdot 1 \cdot 32 : 8 i

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

r = - 4 + 4 i, r = - 4 - 4 i

11 x = 8 x - 6

52 = n - \frac{1}{5} n

4^{9}

\frac{1}{2} (11 + 15)

\frac{3 x - 2}{x + 3} < 6

What are the solutions to the equation r^2+8r+32=0 ?
The solutions to the equation r^2+8r+32=0 are r=-4+4i,r=-4-4i
Find the zeros of r^2+8r+32=0
The zeros of r^2+8r+32=0 are r=-4+4i,r=-4-4i