What are the solutions to the equation 35=k^2-2k ?

The solutions to the equation 35=k^2-2k are k=7,k=-5

Find the zeros of 35=k^2-2k

The zeros of 35=k^2-2k are k=7,k=-5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

35 = k^{2} - 2 k

k = 7, k = - 5

Solution steps

35 = k^{2} - 2 k

Switch sides

k^{2} - 2 k = 35

Move

35

to the left side

k^{2} - 2 k - 35 = 0

Solve with the quadratic formula

k_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot ( - 35 )}{2 \cdot 1}

(- 2)^{2} - 4 \cdot 1 \cdot (- 35) = 12

k_{1, 2} = \frac{- ( - 2 ) \pm 12}{2 \cdot 1}

Separate the solutions

k_{1} = \frac{- ( - 2 ) + 12}{2 \cdot 1}, k_{2} = \frac{- ( - 2 ) - 12}{2 \cdot 1}

k = \frac{- ( - 2 ) + 12}{2 \cdot 1} : 7

k = \frac{- ( - 2 ) - 12}{2 \cdot 1} : - 5

The solutions to the quadratic equation are:

k = 7, k = - 5

563

0 = (x - 2) (x + 5)

3 (2 x - 3) = 4 x

\frac{- 36}{- 6}

24 b^{3} \cdot 40 b^{2} \cdot b^{2}

What are the solutions to the equation 35=k^2-2k ?
The solutions to the equation 35=k^2-2k are k=7,k=-5
Find the zeros of 35=k^2-2k
The zeros of 35=k^2-2k are k=7,k=-5