What are the solutions to the equation 1/5 k(k+4)=0.5 ?

The solutions to the equation 1/5 k(k+4)=0.5 are k=(-4+sqrt(26))/2 ,k=-(4+sqrt(26))/2

Find the zeros of 1/5 k(k+4)=0.5

The zeros of 1/5 k(k+4)=0.5 are k=(-4+sqrt(26))/2 ,k=-(4+sqrt(26))/2

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\frac{1}{5} k (k + 4) = 0.5

k = \frac{- 4 + 26}{2}, k = - \frac{4 + 26}{2}

Decimal Notation

k = 0.54950 \dots, k = - 4.54950 \dots

Solution steps

\frac{1}{5} k (k + 4) = 0.5

Multiply both sides by

10

2 k (k + 4) = 5

Expand

2 k (k + 4) : 2 k^{2} + 8 k

2 k^{2} + 8 k = 5

Move

5

to the left side

2 k^{2} + 8 k - 5 = 0

Solve with the quadratic formula

k_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 2 ( - 5 )}{2 \cdot 2}

8^{2} - 4 \cdot 2 (- 5) = 226

k_{1, 2} = \frac{- 8 \pm 2 26}{2 \cdot 2}

Separate the solutions

k_{1} = \frac{- 8 + 2 26}{2 \cdot 2}, k_{2} = \frac{- 8 - 2 26}{2 \cdot 2}

k = \frac{- 8 + 2 26}{2 \cdot 2} : \frac{- 4 + 26}{2}

k = \frac{- 8 - 2 26}{2 \cdot 2} : - \frac{4 + 26}{2}

The solutions to the quadratic equation are:

k = \frac{- 4 + 26}{2}, k = - \frac{4 + 26}{2}

w^{2} + 4 w = 0

x^{2} - 2 = 8

x^{2} + x - 56 = 0

z, x^{2} + y^{2} = z^{2}

(r^{2} + r + 1) = 0

What are the solutions to the equation 1/5 k(k+4)=0.5 ?
The solutions to the equation 1/5 k(k+4)=0.5 are k=(-4+sqrt(26))/2 ,k=-(4+sqrt(26))/2
Find the zeros of 1/5 k(k+4)=0.5
The zeros of 1/5 k(k+4)=0.5 are k=(-4+sqrt(26))/2 ,k=-(4+sqrt(26))/2