What are the solutions to the equation k^2+7k-5=4 ?

The solutions to the equation k^2+7k-5=4 are k=(-7+sqrt(85))/2 ,k=(-7-sqrt(85))/2

Find the zeros of k^2+7k-5=4

The zeros of k^2+7k-5=4 are k=(-7+sqrt(85))/2 ,k=(-7-sqrt(85))/2

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k^{2} + 7 k - 5 = 4

k = \frac{- 7 + 85}{2}, k = \frac{- 7 - 85}{2}

Decimal Notation

k = 1.10977 \dots, k = - 8.10977 \dots

Solution steps

k^{2} + 7 k - 5 = 4

Move

4

to the left side

k^{2} + 7 k - 9 = 0

Solve with the quadratic formula

k_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 1 \cdot ( - 9 )}{2 \cdot 1}

7^{2} - 4 \cdot 1 \cdot (- 9) = 85

k_{1, 2} = \frac{- 7 \pm 85}{2 \cdot 1}

Separate the solutions

k_{1} = \frac{- 7 + 85}{2 \cdot 1}, k_{2} = \frac{- 7 - 85}{2 \cdot 1}

k = \frac{- 7 + 85}{2 \cdot 1} : \frac{- 7 + 85}{2}

k = \frac{- 7 - 85}{2 \cdot 1} : \frac{- 7 - 85}{2}

The solutions to the quadratic equation are:

k = \frac{- 7 + 85}{2}, k = \frac{- 7 - 85}{2}

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

3 m^{2} - 17 m + 15 = - m

r, V = \frac{π r ^{2} h}{3}

- 3 (x + 6)^{2} - 7 = 0

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

What are the solutions to the equation k^2+7k-5=4 ?
The solutions to the equation k^2+7k-5=4 are k=(-7+sqrt(85))/2 ,k=(-7-sqrt(85))/2
Find the zeros of k^2+7k-5=4
The zeros of k^2+7k-5=4 are k=(-7+sqrt(85))/2 ,k=(-7-sqrt(85))/2