What are the solutions to the equation k^2+16k=-48 ?

The solutions to the equation k^2+16k=-48 are k=-4,k=-12

Find the zeros of k^2+16k=-48

The zeros of k^2+16k=-48 are k=-4,k=-12

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k^{2} + 16 k = - 48

k = - 4, k = - 12

Solution steps

k^{2} + 16 k = - 48

Move

48

to the left side

k^{2} + 16 k + 48 = 0

Solve with the quadratic formula

k_{1, 2} = \frac{- 16 \pm 1 6 ^{2} - 4 \cdot 1 \cdot 48}{2 \cdot 1}

1 6^{2} - 4 \cdot 1 \cdot 48 = 8

k_{1, 2} = \frac{- 16 \pm 8}{2 \cdot 1}

Separate the solutions

k_{1} = \frac{- 16 + 8}{2 \cdot 1}, k_{2} = \frac{- 16 - 8}{2 \cdot 1}

k = \frac{- 16 + 8}{2 \cdot 1} : - 4

k = \frac{- 16 - 8}{2 \cdot 1} : - 12

The solutions to the quadratic equation are:

k = - 4, k = - 12

15 x^{2} - 53 x + 42

a^{3} - 729

0 = \frac{2}{3} (- 4 + 9)^{\frac{3}{2}}

7 a x + 21 x^{2} y

\frac{5}{6} - 1

What are the solutions to the equation k^2+16k=-48 ?
The solutions to the equation k^2+16k=-48 are k=-4,k=-12
Find the zeros of k^2+16k=-48
The zeros of k^2+16k=-48 are k=-4,k=-12