What are the solutions to the equation 1/5 k(k+4)=0.75 ?

The solutions to the equation 1/5 k(k+4)=0.75 are k=(-4+sqrt(31))/2 ,k=-(4+sqrt(31))/2

Find the zeros of 1/5 k(k+4)=0.75

The zeros of 1/5 k(k+4)=0.75 are k=(-4+sqrt(31))/2 ,k=-(4+sqrt(31))/2

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\frac{1}{5} k (k + 4) = 0.75

k = \frac{- 4 + 31}{2}, k = - \frac{4 + 31}{2}

Decimal Notation

k = 0.78388 \dots, k = - 4.78388 \dots

Solution steps

\frac{1}{5} k (k + 4) = 0.75

Multiply both sides by

100

20 k (k + 4) = 75

Expand

20 k (k + 4) : 20 k^{2} + 80 k

20 k^{2} + 80 k = 75

Move

75

to the left side

20 k^{2} + 80 k - 75 = 0

Solve with the quadratic formula

k_{1, 2} = \frac{- 80 \pm 8 0 ^{2} - 4 \cdot 20 ( - 75 )}{2 \cdot 20}

8 0^{2} - 4 \cdot 20 (- 75) = 2031

k_{1, 2} = \frac{- 80 \pm 20 31}{2 \cdot 20}

Separate the solutions

k_{1} = \frac{- 80 + 20 31}{2 \cdot 20}, k_{2} = \frac{- 80 - 20 31}{2 \cdot 20}

k = \frac{- 80 + 20 31}{2 \cdot 20} : \frac{- 4 + 31}{2}

k = \frac{- 80 - 20 31}{2 \cdot 20} : - \frac{4 + 31}{2}

The solutions to the quadratic equation are:

k = \frac{- 4 + 31}{2}, k = - \frac{4 + 31}{2}

(x - 1) (x + 9) = 0

x^{2} + 2 x^{2} = 1

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

144 a^{2} + 216 a + 81 = 0

(x - 1) (2 x + 2) = 0

What are the solutions to the equation 1/5 k(k+4)=0.75 ?
The solutions to the equation 1/5 k(k+4)=0.75 are k=(-4+sqrt(31))/2 ,k=-(4+sqrt(31))/2
Find the zeros of 1/5 k(k+4)=0.75
The zeros of 1/5 k(k+4)=0.75 are k=(-4+sqrt(31))/2 ,k=-(4+sqrt(31))/2