What are the solutions to the equation k^2-4k-21=0 ?

The solutions to the equation k^2-4k-21=0 are k=7,k=-3

Find the zeros of k^2-4k-21=0

The zeros of k^2-4k-21=0 are k=7,k=-3

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k^{2} - 4 k - 21 = 0

k = 7, k = - 3

Solution steps

k^{2} - 4 k - 21 = 0

Solve with the quadratic formula

k_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot ( - 21 )}{2 \cdot 1}

(- 4)^{2} - 4 \cdot 1 \cdot (- 21) = 10

k_{1, 2} = \frac{- ( - 4 ) \pm 10}{2 \cdot 1}

Separate the solutions

k_{1} = \frac{- ( - 4 ) + 10}{2 \cdot 1}, k_{2} = \frac{- ( - 4 ) - 10}{2 \cdot 1}

k = \frac{- ( - 4 ) + 10}{2 \cdot 1} : 7

k = \frac{- ( - 4 ) - 10}{2 \cdot 1} : - 3

The solutions to the quadratic equation are:

k = 7, k = - 3

(x^{2} + 4 x + 4) = 0

lo g_{10} (3) x^{2} - lo g_{10} (3) = 2

8 x - x^{2}

x^{2} + 6 x - 17 = 0

b, m = 4 f b^{2} - g^{2}

What are the solutions to the equation k^2-4k-21=0 ?
The solutions to the equation k^2-4k-21=0 are k=7,k=-3
Find the zeros of k^2-4k-21=0
The zeros of k^2-4k-21=0 are k=7,k=-3