What are the solutions to the equation 2k^2+2=4k ?

The solutions to the equation 2k^2+2=4k are k=1

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2k^2+2=4k

2 k^{2} + 2 = 4 k

k = 1

Solution steps

2 k^{2} + 2 = 4 k

Move

4 k

to the left side

2 k^{2} + 2 - 4 k = 0

Write in the standard form

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

2 k^{2} - 4 k + 2 = 0

Solve with the quadratic formula

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

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

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

k = \frac{- ( - 4 )}{2 \cdot 2}

\frac{- ( - 4 )}{2 \cdot 2} = 1

k = 1

The solution to the quadratic equation is:

k = 1

- \frac{7}{2} \cdot \frac{1}{6}

(\frac{1}{2})^{2} + (\frac{1}{4})^{2}

7^{p + 2} \leq 1 3^{5 - p}

4 \cdot \frac{4}{3}

v^{2} - 9 v + 8 = 0

What are the solutions to the equation 2k^2+2=4k ?
The solutions to the equation 2k^2+2=4k are k=1
Find the zeros of 2k^2+2=4k
The zeros of 2k^2+2=4k are k=1