What are the solutions to the equation 5k^2+28=27k ?

The solutions to the equation 5k^2+28=27k are k=4,k= 7/5

Find the zeros of 5k^2+28=27k

The zeros of 5k^2+28=27k are k=4,k= 7/5

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5 k^{2} + 28 = 27 k

k = 4, k = \frac{7}{5}

Decimal Notation

k = 4, k = 1.4

Solution steps

5 k^{2} + 28 = 27 k

Move

27 k

to the left side

5 k^{2} + 28 - 27 k = 0

Write in the standard form

a x^{2} + b x + c = 0

5 k^{2} - 27 k + 28 = 0

Solve with the quadratic formula

k_{1, 2} = \frac{- ( - 27 ) \pm ( - 27 ) ^{2} - 4 \cdot 5 \cdot 28}{2 \cdot 5}

(- 27)^{2} - 4 \cdot 5 \cdot 28 = 13

k_{1, 2} = \frac{- ( - 27 ) \pm 13}{2 \cdot 5}

Separate the solutions

k_{1} = \frac{- ( - 27 ) + 13}{2 \cdot 5}, k_{2} = \frac{- ( - 27 ) - 13}{2 \cdot 5}

k = \frac{- ( - 27 ) + 13}{2 \cdot 5} : 4

k = \frac{- ( - 27 ) - 13}{2 \cdot 5} : \frac{7}{5}

The solutions to the quadratic equation are:

k = 4, k = \frac{7}{5}

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What are the solutions to the equation 5k^2+28=27k ?
The solutions to the equation 5k^2+28=27k are k=4,k= 7/5
Find the zeros of 5k^2+28=27k
The zeros of 5k^2+28=27k are k=4,k= 7/5