What are the solutions to the equation 6r^2+8=-4r ?

The solutions to the equation 6r^2+8=-4r are r=-1/3+i(sqrt(11))/3 ,r=-1/3-i(sqrt(11))/3

Find the zeros of 6r^2+8=-4r

The zeros of 6r^2+8=-4r are r=-1/3+i(sqrt(11))/3 ,r=-1/3-i(sqrt(11))/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

6 r^{2} + 8 = - 4 r

r = - \frac{1}{3} + i \frac{11}{3}, r = - \frac{1}{3} - i \frac{11}{3}

Solution steps

6 r^{2} + 8 = - 4 r

Move

4 r

to the left side

6 r^{2} + 8 + 4 r = 0

Write in the standard form

a x^{2} + b x + c = 0

6 r^{2} + 4 r + 8 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 6 \cdot 8}{2 \cdot 6}

Simplify

4^{2} - 4 \cdot 6 \cdot 8 : 411 i

r_{1, 2} = \frac{- 4 \pm 4 11 i}{2 \cdot 6}

Separate the solutions

r_{1} = \frac{- 4 + 4 11 i}{2 \cdot 6}, r_{2} = \frac{- 4 - 4 11 i}{2 \cdot 6}

r = \frac{- 4 + 4 11 i}{2 \cdot 6} : - \frac{1}{3} + i \frac{11}{3}

r = \frac{- 4 - 4 11 i}{2 \cdot 6} : - \frac{1}{3} - i \frac{11}{3}

The solutions to the quadratic equation are:

r = - \frac{1}{3} + i \frac{11}{3}, r = - \frac{1}{3} - i \frac{11}{3}

9 x^{2} - 3 x = 20

- x^{2} + x - 1 = 0

x^{2} + 8 x - 23 = 0

x^{2} - 4 x + 8 = 0

4 y^{2} + 8 y + 5 = 0

What are the solutions to the equation 6r^2+8=-4r ?
The solutions to the equation 6r^2+8=-4r are r=-1/3+i(sqrt(11))/3 ,r=-1/3-i(sqrt(11))/3
Find the zeros of 6r^2+8=-4r
The zeros of 6r^2+8=-4r are r=-1/3+i(sqrt(11))/3 ,r=-1/3-i(sqrt(11))/3