What are the solutions to the equation 35k^2-22k+7=4 ?

The solutions to the equation 35k^2-22k+7=4 are k= 3/7 ,k= 1/5

Find the zeros of 35k^2-22k+7=4

The zeros of 35k^2-22k+7=4 are k= 3/7 ,k= 1/5

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35 k^{2} - 22 k + 7 = 4

k = \frac{3}{7}, k = \frac{1}{5}

Decimal Notation

k = 0.42857 \dots, k = 0.2

Solution steps

35 k^{2} - 22 k + 7 = 4

Move

4

to the left side

35 k^{2} - 22 k + 3 = 0

Solve with the quadratic formula

k_{1, 2} = \frac{- ( - 22 ) \pm ( - 22 ) ^{2} - 4 \cdot 35 \cdot 3}{2 \cdot 35}

(- 22)^{2} - 4 \cdot 35 \cdot 3 = 8

k_{1, 2} = \frac{- ( - 22 ) \pm 8}{2 \cdot 35}

Separate the solutions

k_{1} = \frac{- ( - 22 ) + 8}{2 \cdot 35}, k_{2} = \frac{- ( - 22 ) - 8}{2 \cdot 35}

k = \frac{- ( - 22 ) + 8}{2 \cdot 35} : \frac{3}{7}

k = \frac{- ( - 22 ) - 8}{2 \cdot 35} : \frac{1}{5}

The solutions to the quadratic equation are:

k = \frac{3}{7}, k = \frac{1}{5}

\frac{x + 2}{2} + 2 = \frac{x + 3}{5}

10 = - 4.9 t^{2} + 18.01 t

2 w^{4} - 7 w^{3} - 15 w^{2}

\frac{- 3}{5} + \frac{- 1}{15} + \frac{8}{25}

(x^{2} - 1)^{18} (x^{2} - x)^{17} \geq 0

What are the solutions to the equation 35k^2-22k+7=4 ?
The solutions to the equation 35k^2-22k+7=4 are k= 3/7 ,k= 1/5
Find the zeros of 35k^2-22k+7=4
The zeros of 35k^2-22k+7=4 are k= 3/7 ,k= 1/5