What are the solutions to the equation k^2+16k+42=28 ?

The solutions to the equation k^2+16k+42=28 are k=-8+5sqrt(2),k=-8-5sqrt(2)

Find the zeros of k^2+16k+42=28

The zeros of k^2+16k+42=28 are k=-8+5sqrt(2),k=-8-5sqrt(2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

k^{2} + 16 k + 42 = 28

k = - 8 + 52, k = - 8 - 52

Decimal Notation

k = - 0.92893 \dots, k = - 15.07106 \dots

Solution steps

k^{2} + 16 k + 42 = 28

Move

28

to the left side

k^{2} + 16 k + 14 = 0

Solve with the quadratic formula

k_{1, 2} = \frac{- 16 \pm 1 6 ^{2} - 4 \cdot 1 \cdot 14}{2 \cdot 1}

1 6^{2} - 4 \cdot 1 \cdot 14 = 102

k_{1, 2} = \frac{- 16 \pm 10 2}{2 \cdot 1}

Separate the solutions

k_{1} = \frac{- 16 + 10 2}{2 \cdot 1}, k_{2} = \frac{- 16 - 10 2}{2 \cdot 1}

k = \frac{- 16 + 10 2}{2 \cdot 1} : - 8 + 52

k = \frac{- 16 - 10 2}{2 \cdot 1} : - 8 - 52

The solutions to the quadratic equation are:

k = - 8 + 52, k = - 8 - 52

- 4 x \geq - 12

x^{2} + 6.3 \cdot 1 0^{- 5} x - 3.4 \cdot 1 0^{- 6} = 0

(\frac{x}{2}) - 9 < \frac{( 1 - 2 x )}{3}

x^{5} - 9 x^{3}

- 0.54395 x^{2} - 1 0^{- 9.7} x + 1 0^{- 9.7} = 0

What are the solutions to the equation k^2+16k+42=28 ?
The solutions to the equation k^2+16k+42=28 are k=-8+5sqrt(2),k=-8-5sqrt(2)
Find the zeros of k^2+16k+42=28
The zeros of k^2+16k+42=28 are k=-8+5sqrt(2),k=-8-5sqrt(2)