What are the solutions to the equation 4k(k+3)=-11 ?

The solutions to the equation 4k(k+3)=-11 are k=-3/2+i(sqrt(2))/2 ,k=-3/2-i(sqrt(2))/2

Find the zeros of 4k(k+3)=-11

The zeros of 4k(k+3)=-11 are k=-3/2+i(sqrt(2))/2 ,k=-3/2-i(sqrt(2))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4 k (k + 3) = - 11

k = - \frac{3}{2} + i \frac{2}{2}, k = - \frac{3}{2} - i \frac{2}{2}

Solution steps

4 k (k + 3) = - 11

Expand

4 k (k + 3) : 4 k^{2} + 12 k

4 k^{2} + 12 k = - 11

Move

11

to the left side

4 k^{2} + 12 k + 11 = 0

Solve with the quadratic formula

k_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 4 \cdot 11}{2 \cdot 4}

Simplify

1 2^{2} - 4 \cdot 4 \cdot 11 : 42 i

k_{1, 2} = \frac{- 12 \pm 4 2 i}{2 \cdot 4}

Separate the solutions

k_{1} = \frac{- 12 + 4 2 i}{2 \cdot 4}, k_{2} = \frac{- 12 - 4 2 i}{2 \cdot 4}

k = \frac{- 12 + 4 2 i}{2 \cdot 4} : - \frac{3}{2} + i \frac{2}{2}

k = \frac{- 12 - 4 2 i}{2 \cdot 4} : - \frac{3}{2} - i \frac{2}{2}

The solutions to the quadratic equation are:

k = - \frac{3}{2} + i \frac{2}{2}, k = - \frac{3}{2} - i \frac{2}{2}

3 x^{2} - 3 x = 9

2 p^{2} + 8 p + (k + 1) = 0

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

s^{2} + 2 s + 2 = 0

\frac{x}{3} = x^{2}

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