What are the solutions to the equation 2k^2-13k+12=0 ?

The solutions to the equation 2k^2-13k+12=0 are k=(13+sqrt(73))/4 ,k=(13-sqrt(73))/4

Find the zeros of 2k^2-13k+12=0

The zeros of 2k^2-13k+12=0 are k=(13+sqrt(73))/4 ,k=(13-sqrt(73))/4

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2 k^{2} - 13 k + 12 = 0

k = \frac{13 + 73}{4}, k = \frac{13 - 73}{4}

Decimal Notation

k = 5.38600 \dots, k = 1.11399 \dots

Solution steps

2 k^{2} - 13 k + 12 = 0

Solve with the quadratic formula

k_{1, 2} = \frac{- ( - 13 ) \pm ( - 13 ) ^{2} - 4 \cdot 2 \cdot 12}{2 \cdot 2}

(- 13)^{2} - 4 \cdot 2 \cdot 12 = 73

k_{1, 2} = \frac{- ( - 13 ) \pm 73}{2 \cdot 2}

Separate the solutions

k_{1} = \frac{- ( - 13 ) + 73}{2 \cdot 2}, k_{2} = \frac{- ( - 13 ) - 73}{2 \cdot 2}

k = \frac{- ( - 13 ) + 73}{2 \cdot 2} : \frac{13 + 73}{4}

k = \frac{- ( - 13 ) - 73}{2 \cdot 2} : \frac{13 - 73}{4}

The solutions to the quadratic equation are:

k = \frac{13 + 73}{4}, k = \frac{13 - 73}{4}

\frac{8}{9} - \frac{5}{6} (\frac{1}{3})

∣ 6 - 2 x ∣ + 1 = 3

- 2 x > 4

4 x^{2} + 6 = - 11 x

- 3 x^{2} - 5 x + 1 = 0

What are the solutions to the equation 2k^2-13k+12=0 ?
The solutions to the equation 2k^2-13k+12=0 are k=(13+sqrt(73))/4 ,k=(13-sqrt(73))/4
Find the zeros of 2k^2-13k+12=0
The zeros of 2k^2-13k+12=0 are k=(13+sqrt(73))/4 ,k=(13-sqrt(73))/4