What are the solutions to the equation 2k^2+k-6=0 ?

The solutions to the equation 2k^2+k-6=0 are k= 3/2 ,k=-2

Find the zeros of 2k^2+k-6=0

The zeros of 2k^2+k-6=0 are k= 3/2 ,k=-2

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

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

Decimal Notation

k = 1.5, k = - 2

Solution steps

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

Solve with the quadratic formula

k_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 2 ( - 6 )}{2 \cdot 2}

1^{2} - 4 \cdot 2 (- 6) = 7

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

Separate the solutions

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

k = \frac{- 1 + 7}{2 \cdot 2} : \frac{3}{2}

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

The solutions to the quadratic equation are:

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

\frac{2}{7} - \frac{3}{2 b}

x^{2} - 3.2 x + 2.56 - 0.0909090909 = 0

5 x^{2} + 8 x \leq 4

(9 x^{10} y^{3}) (- x^{5} y^{3})

- 3 x - 4 > 2

What are the solutions to the equation 2k^2+k-6=0 ?
The solutions to the equation 2k^2+k-6=0 are k= 3/2 ,k=-2
Find the zeros of 2k^2+k-6=0
The zeros of 2k^2+k-6=0 are k= 3/2 ,k=-2