What are the solutions to the equation 4k(k+13)=13 ?

The solutions to the equation 4k(k+13)=13 are k=(-13+sqrt(182))/2 ,k=-(13+sqrt(182))/2

Find the zeros of 4k(k+13)=13

The zeros of 4k(k+13)=13 are k=(-13+sqrt(182))/2 ,k=-(13+sqrt(182))/2

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4 k (k + 13) = 13

k = \frac{- 13 + 182}{2}, k = - \frac{13 + 182}{2}

Decimal Notation

k = 0.24536 \dots, k = - 13.24536 \dots

Solution steps

4 k (k + 13) = 13

Expand

4 k (k + 13) : 4 k^{2} + 52 k

4 k^{2} + 52 k = 13

Move

13

to the left side

4 k^{2} + 52 k - 13 = 0

Solve with the quadratic formula

k_{1, 2} = \frac{- 52 \pm 5 2 ^{2} - 4 \cdot 4 ( - 13 )}{2 \cdot 4}

5 2^{2} - 4 \cdot 4 (- 13) = 4182

k_{1, 2} = \frac{- 52 \pm 4 182}{2 \cdot 4}

Separate the solutions

k_{1} = \frac{- 52 + 4 182}{2 \cdot 4}, k_{2} = \frac{- 52 - 4 182}{2 \cdot 4}

k = \frac{- 52 + 4 182}{2 \cdot 4} : \frac{- 13 + 182}{2}

k = \frac{- 52 - 4 182}{2 \cdot 4} : - \frac{13 + 182}{2}

The solutions to the quadratic equation are:

k = \frac{- 13 + 182}{2}, k = - \frac{13 + 182}{2}

7 x^{2} + x = 0

x^{2} + 22 x + 40 = 0

2 x^{2} + 12 x - 6 = 0

x^{2} = 12 \cdot 3

3 x^{2} + 135 = 0

What are the solutions to the equation 4k(k+13)=13 ?
The solutions to the equation 4k(k+13)=13 are k=(-13+sqrt(182))/2 ,k=-(13+sqrt(182))/2
Find the zeros of 4k(k+13)=13
The zeros of 4k(k+13)=13 are k=(-13+sqrt(182))/2 ,k=-(13+sqrt(182))/2