What are the solutions to the equation 3k^2+1k=-10 ?

The solutions to the equation 3k^2+1k=-10 are k=-1/6+i(sqrt(119))/6 ,k=-1/6-i(sqrt(119))/6

Find the zeros of 3k^2+1k=-10

The zeros of 3k^2+1k=-10 are k=-1/6+i(sqrt(119))/6 ,k=-1/6-i(sqrt(119))/6

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

3 k^{2} + 1 k = - 10

k = - \frac{1}{6} + i \frac{119}{6}, k = - \frac{1}{6} - i \frac{119}{6}

Solution steps

3 k^{2} + 1 \cdot k = - 10

Refine

3 k^{2} + k = - 10

Move

10

to the left side

3 k^{2} + k + 10 = 0

Solve with the quadratic formula

k_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 3 \cdot 10}{2 \cdot 3}

Simplify

1^{2} - 4 \cdot 3 \cdot 10 : 119 i

k_{1, 2} = \frac{- 1 \pm 119 i}{2 \cdot 3}

Separate the solutions

k_{1} = \frac{- 1 + 119 i}{2 \cdot 3}, k_{2} = \frac{- 1 - 119 i}{2 \cdot 3}

k = \frac{- 1 + 119 i}{2 \cdot 3} : - \frac{1}{6} + i \frac{119}{6}

k = \frac{- 1 - 119 i}{2 \cdot 3} : - \frac{1}{6} - i \frac{119}{6}

The solutions to the quadratic equation are:

k = - \frac{1}{6} + i \frac{119}{6}, k = - \frac{1}{6} - i \frac{119}{6}

\frac{( ( x + h ) ^{(\frac{3}{2})} - x ^{\frac{3}{2}} )}{h}

- 4 x + 5 = - 4 x - 5

18 \cdot \frac{1}{3}

∣ (x + 2) - 5 ∣ \geq \frac{1}{2} x > 5

lo g_{7} (a) \cdot lo g_{b} (7)

What are the solutions to the equation 3k^2+1k=-10 ?
The solutions to the equation 3k^2+1k=-10 are k=-1/6+i(sqrt(119))/6 ,k=-1/6-i(sqrt(119))/6
Find the zeros of 3k^2+1k=-10
The zeros of 3k^2+1k=-10 are k=-1/6+i(sqrt(119))/6 ,k=-1/6-i(sqrt(119))/6