What are the solutions to the equation k^2-15k=-36 ?

The solutions to the equation k^2-15k=-36 are k=12,k=3

Find the zeros of k^2-15k=-36

The zeros of k^2-15k=-36 are k=12,k=3

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k^{2} - 15 k = - 36

k = 12, k = 3

Solution steps

k^{2} - 15 k = - 36

Move

36

to the left side

k^{2} - 15 k + 36 = 0

Solve with the quadratic formula

k_{1, 2} = \frac{- ( - 15 ) \pm ( - 15 ) ^{2} - 4 \cdot 1 \cdot 36}{2 \cdot 1}

(- 15)^{2} - 4 \cdot 1 \cdot 36 = 9

k_{1, 2} = \frac{- ( - 15 ) \pm 9}{2 \cdot 1}

Separate the solutions

k_{1} = \frac{- ( - 15 ) + 9}{2 \cdot 1}, k_{2} = \frac{- ( - 15 ) - 9}{2 \cdot 1}

k = \frac{- ( - 15 ) + 9}{2 \cdot 1} : 12

k = \frac{- ( - 15 ) - 9}{2 \cdot 1} : 3

The solutions to the quadratic equation are:

k = 12, k = 3

e^{- 2 l n (x)}

2 (- 2) + 4

9 x^{2} - 6 x + 2 = 0

100 x^{2} - 81

\frac{5 π}{12} - π

What are the solutions to the equation k^2-15k=-36 ?
The solutions to the equation k^2-15k=-36 are k=12,k=3
Find the zeros of k^2-15k=-36
The zeros of k^2-15k=-36 are k=12,k=3