What are the solutions to the equation k^2+15k=-56 ?

The solutions to the equation k^2+15k=-56 are k=-7,k=-8

Find the zeros of k^2+15k=-56

The zeros of k^2+15k=-56 are k=-7,k=-8

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k^{2} + 15 k = - 56

k = - 7, k = - 8

Solution steps

k^{2} + 15 k = - 56

Move

56

to the left side

k^{2} + 15 k + 56 = 0

Solve with the quadratic formula

k_{1, 2} = \frac{- 15 \pm 1 5 ^{2} - 4 \cdot 1 \cdot 56}{2 \cdot 1}

1 5^{2} - 4 \cdot 1 \cdot 56 = 1

k_{1, 2} = \frac{- 15 \pm 1}{2 \cdot 1}

Separate the solutions

k_{1} = \frac{- 15 + 1}{2 \cdot 1}, k_{2} = \frac{- 15 - 1}{2 \cdot 1}

k = \frac{- 15 + 1}{2 \cdot 1} : - 7

k = \frac{- 15 - 1}{2 \cdot 1} : - 8

The solutions to the quadratic equation are:

k = - 7, k = - 8

(2 - 3) (6 + 5)

6 x^{2} - 14 x - 98 = 0

\frac{\frac{7}{4} - \frac{2}{3} ( \frac{7}{2} + 1 )}{24 + 2 ( 3 + 2 )}

\frac{( x - 5 ) ^{2}}{x ^{2} - 16} \geq 0

(x - 7)^{2} = 25

What are the solutions to the equation k^2+15k=-56 ?
The solutions to the equation k^2+15k=-56 are k=-7,k=-8
Find the zeros of k^2+15k=-56
The zeros of k^2+15k=-56 are k=-7,k=-8