What are the solutions to the equation 4k^2-6k-1=0 ?

The solutions to the equation 4k^2-6k-1=0 are k=(3+sqrt(13))/4 ,k=(3-sqrt(13))/4

Find the zeros of 4k^2-6k-1=0

The zeros of 4k^2-6k-1=0 are k=(3+sqrt(13))/4 ,k=(3-sqrt(13))/4

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4 k^{2} - 6 k - 1 = 0

k = \frac{3 + 13}{4}, k = \frac{3 - 13}{4}

Decimal Notation

k = 1.65138 \dots, k = - 0.15138 \dots

Solution steps

4 k^{2} - 6 k - 1 = 0

Solve with the quadratic formula

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

(- 6)^{2} - 4 \cdot 4 (- 1) = 213

k_{1, 2} = \frac{- ( - 6 ) \pm 2 13}{2 \cdot 4}

Separate the solutions

k_{1} = \frac{- ( - 6 ) + 2 13}{2 \cdot 4}, k_{2} = \frac{- ( - 6 ) - 2 13}{2 \cdot 4}

k = \frac{- ( - 6 ) + 2 13}{2 \cdot 4} : \frac{3 + 13}{4}

k = \frac{- ( - 6 ) - 2 13}{2 \cdot 4} : \frac{3 - 13}{4}

The solutions to the quadratic equation are:

k = \frac{3 + 13}{4}, k = \frac{3 - 13}{4}

\frac{- 120}{5}

\frac{2 + 3 i}{5 + 2 i}

\frac{x + 4}{5} = \frac{x - 3}{4}

a^{2} + a^{3} + a^{4}

(\frac{2 ^{\frac{3}{4}}}{2 ^{\frac{1}{4}}})^{6}

What are the solutions to the equation 4k^2-6k-1=0 ?
The solutions to the equation 4k^2-6k-1=0 are k=(3+sqrt(13))/4 ,k=(3-sqrt(13))/4
Find the zeros of 4k^2-6k-1=0
The zeros of 4k^2-6k-1=0 are k=(3+sqrt(13))/4 ,k=(3-sqrt(13))/4