What are the solutions to the equation r^2+4r+5=0 ?

The solutions to the equation r^2+4r+5=0 are r=-2+i,r=-2-i

Find the zeros of r^2+4r+5=0

The zeros of r^2+4r+5=0 are r=-2+i,r=-2-i

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

r^{2} + 4 r + 5 = 0

r = - 2 + i, r = - 2 - i

Solution steps

r^{2} + 4 r + 5 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot 5}{2 \cdot 1}

Simplify

4^{2} - 4 \cdot 1 \cdot 5 : 2 i

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

r = - 2 + i, r = - 2 - i

4 x^{2} + 12 x = 8 x - 4

x (x + 4) = 117

5 - 8 v < 11 - 6 v

4 x^{2} - 12 x - 27 = 0

2 k^{2} + 13 k + 11

What are the solutions to the equation r^2+4r+5=0 ?
The solutions to the equation r^2+4r+5=0 are r=-2+i,r=-2-i
Find the zeros of r^2+4r+5=0
The zeros of r^2+4r+5=0 are r=-2+i,r=-2-i