What are the solutions to the equation k(2k+5)=3 ?

The solutions to the equation k(2k+5)=3 are k= 1/2 ,k=-3

Find the zeros of k(2k+5)=3

The zeros of k(2k+5)=3 are k= 1/2 ,k=-3

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k (2 k + 5) = 3

k = \frac{1}{2}, k = - 3

Decimal Notation

k = 0.5, k = - 3

Solution steps

k (2 k + 5) = 3

Expand

k (2 k + 5) : 2 k^{2} + 5 k

2 k^{2} + 5 k = 3

Move

3

to the left side

2 k^{2} + 5 k - 3 = 0

Solve with the quadratic formula

k_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 2 ( - 3 )}{2 \cdot 2}

5^{2} - 4 \cdot 2 (- 3) = 7

k_{1, 2} = \frac{- 5 \pm 7}{2 \cdot 2}

Separate the solutions

k_{1} = \frac{- 5 + 7}{2 \cdot 2}, k_{2} = \frac{- 5 - 7}{2 \cdot 2}

k = \frac{- 5 + 7}{2 \cdot 2} : \frac{1}{2}

k = \frac{- 5 - 7}{2 \cdot 2} : - 3

The solutions to the quadratic equation are:

k = \frac{1}{2}, k = - 3

- 10 x - 47 = - x^{2} + 1

x, (x + y)^{2} = 20

5 x^{2} - 8 = 0

3 x^{2} + 12 x + 6 = 6

(x + 2)^{2} = 4 x + 20

What are the solutions to the equation k(2k+5)=3 ?
The solutions to the equation k(2k+5)=3 are k= 1/2 ,k=-3
Find the zeros of k(2k+5)=3
The zeros of k(2k+5)=3 are k= 1/2 ,k=-3