What are the solutions to the equation 21k^2+20k-32=-7 ?

The solutions to the equation 21k^2+20k-32=-7 are k= 5/7 ,k=-5/3

Find the zeros of 21k^2+20k-32=-7

The zeros of 21k^2+20k-32=-7 are k= 5/7 ,k=-5/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

21 k^{2} + 20 k - 32 = - 7

k = \frac{5}{7}, k = - \frac{5}{3}

Decimal Notation

k = 0.71428 \dots, k = - 1.66666 \dots

Solution steps

21 k^{2} + 20 k - 32 = - 7

Move

7

to the left side

21 k^{2} + 20 k - 25 = 0

Solve with the quadratic formula

k_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 \cdot 21 ( - 25 )}{2 \cdot 21}

2 0^{2} - 4 \cdot 21 (- 25) = 50

k_{1, 2} = \frac{- 20 \pm 50}{2 \cdot 21}

Separate the solutions

k_{1} = \frac{- 20 + 50}{2 \cdot 21}, k_{2} = \frac{- 20 - 50}{2 \cdot 21}

k = \frac{- 20 + 50}{2 \cdot 21} : \frac{5}{7}

k = \frac{- 20 - 50}{2 \cdot 21} : - \frac{5}{3}

The solutions to the quadratic equation are:

k = \frac{5}{7}, k = - \frac{5}{3}

\frac{1}{6} x - \frac{1}{2} (x - 1) = \frac{1}{2}

2 x^{2} = 6 x + 3

6 y^{2} - 12 y = 0

8 x^{6} - 27 y^{9}

\frac{2 x + 1}{x + 1} \leq 2

What are the solutions to the equation 21k^2+20k-32=-7 ?
The solutions to the equation 21k^2+20k-32=-7 are k= 5/7 ,k=-5/3
Find the zeros of 21k^2+20k-32=-7
The zeros of 21k^2+20k-32=-7 are k= 5/7 ,k=-5/3