What are the solutions to the equation r^2+12r+37=0 ?

The solutions to the equation r^2+12r+37=0 are r=-6+i,r=-6-i

Find the zeros of r^2+12r+37=0

The zeros of r^2+12r+37=0 are r=-6+i,r=-6-i

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

r^{2} + 12 r + 37 = 0

r = - 6 + i, r = - 6 - i

Solution steps

r^{2} + 12 r + 37 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 1 \cdot 37}{2 \cdot 1}

Simplify

1 2^{2} - 4 \cdot 1 \cdot 37 : 2 i

r_{1, 2} = \frac{- 12 \pm 2 i}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- 12 + 2 i}{2 \cdot 1}, r_{2} = \frac{- 12 - 2 i}{2 \cdot 1}

r = \frac{- 12 + 2 i}{2 \cdot 1} : - 6 + i

r = \frac{- 12 - 2 i}{2 \cdot 1} : - 6 - i

The solutions to the quadratic equation are:

r = - 6 + i, r = - 6 - i

5 x^{2} + 9 x = 4

3 x^{3} - 2 x^{2} + 12 x - 8

15 x - 21

x^{(2)} + 2 x - 3 = 0

x (x^{\frac{1}{3}})

What are the solutions to the equation r^2+12r+37=0 ?
The solutions to the equation r^2+12r+37=0 are r=-6+i,r=-6-i
Find the zeros of r^2+12r+37=0
The zeros of r^2+12r+37=0 are r=-6+i,r=-6-i