What are the solutions to the equation 4k^2-4k-4=-5k-5 ?

The solutions to the equation 4k^2-4k-4=-5k-5 are k=-1/8+i(sqrt(15))/8 ,k=-1/8-i(sqrt(15))/8

Find the zeros of 4k^2-4k-4=-5k-5

The zeros of 4k^2-4k-4=-5k-5 are k=-1/8+i(sqrt(15))/8 ,k=-1/8-i(sqrt(15))/8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4 k^{2} - 4 k - 4 = - 5 k - 5

k = - \frac{1}{8} + i \frac{15}{8}, k = - \frac{1}{8} - i \frac{15}{8}

Solution steps

4 k^{2} - 4 k - 4 = - 5 k - 5

Move

5

to the left side

4 k^{2} - 4 k + 1 = - 5 k

Move

5 k

to the left side

4 k^{2} + k + 1 = 0

Solve with the quadratic formula

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

Simplify

1^{2} - 4 \cdot 4 \cdot 1 : 15 i

k_{1, 2} = \frac{- 1 \pm 15 i}{2 \cdot 4}

Separate the solutions

k_{1} = \frac{- 1 + 15 i}{2 \cdot 4}, k_{2} = \frac{- 1 - 15 i}{2 \cdot 4}

k = \frac{- 1 + 15 i}{2 \cdot 4} : - \frac{1}{8} + i \frac{15}{8}

k = \frac{- 1 - 15 i}{2 \cdot 4} : - \frac{1}{8} - i \frac{15}{8}

The solutions to the quadratic equation are:

k = - \frac{1}{8} + i \frac{15}{8}, k = - \frac{1}{8} - i \frac{15}{8}

x^{2} - 1 = 399

x^{2} + 9 x + 30 = 0

a^{2} = 144

a^{2} = 100

(m^{2} + 2 m - 1) = 0

What are the solutions to the equation 4k^2-4k-4=-5k-5 ?
The solutions to the equation 4k^2-4k-4=-5k-5 are k=-1/8+i(sqrt(15))/8 ,k=-1/8-i(sqrt(15))/8
Find the zeros of 4k^2-4k-4=-5k-5
The zeros of 4k^2-4k-4=-5k-5 are k=-1/8+i(sqrt(15))/8 ,k=-1/8-i(sqrt(15))/8