What are the solutions to the equation 48+2k=k^2 ?

The solutions to the equation 48+2k=k^2 are k=8,k=-6

Find the zeros of 48+2k=k^2

The zeros of 48+2k=k^2 are k=8,k=-6

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48 + 2 k = k^{2}

k = 8, k = - 6

Solution steps

48 + 2 k = k^{2}

Switch sides

k^{2} = 48 + 2 k

Move

2 k

to the left side

k^{2} - 2 k = 48

Move

48

to the left side

k^{2} - 2 k - 48 = 0

Solve with the quadratic formula

k_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot ( - 48 )}{2 \cdot 1}

(- 2)^{2} - 4 \cdot 1 \cdot (- 48) = 14

k_{1, 2} = \frac{- ( - 2 ) \pm 14}{2 \cdot 1}

Separate the solutions

k_{1} = \frac{- ( - 2 ) + 14}{2 \cdot 1}, k_{2} = \frac{- ( - 2 ) - 14}{2 \cdot 1}

k = \frac{- ( - 2 ) + 14}{2 \cdot 1} : 8

k = \frac{- ( - 2 ) - 14}{2 \cdot 1} : - 6

The solutions to the quadratic equation are:

k = 8, k = - 6

x^{2} + 26 x + 168

∣ 6 u + 24 ∣ = 0

\frac{2}{3} x = 16

2 x + 4 = \frac{6 ( 2 x + 10 )}{3}

What are the solutions to the equation 48+2k=k^2 ?
The solutions to the equation 48+2k=k^2 are k=8,k=-6
Find the zeros of 48+2k=k^2
The zeros of 48+2k=k^2 are k=8,k=-6