What are the solutions to the equation -k^2-3k-2=-5k^2 ?

The solutions to the equation -k^2-3k-2=-5k^2 are k=(3+sqrt(41))/8 ,k=(3-sqrt(41))/8

Find the zeros of -k^2-3k-2=-5k^2

The zeros of -k^2-3k-2=-5k^2 are k=(3+sqrt(41))/8 ,k=(3-sqrt(41))/8

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

k = \frac{3 + 41}{8}, k = \frac{3 - 41}{8}

Decimal Notation

k = 1.17539 \dots, k = - 0.42539 \dots

Solution steps

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

Move

5 k^{2}

to the left side

4 k^{2} - 3 k - 2 = 0

Solve with the quadratic formula

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

(- 3)^{2} - 4 \cdot 4 (- 2) = 41

k_{1, 2} = \frac{- ( - 3 ) \pm 41}{2 \cdot 4}

Separate the solutions

k_{1} = \frac{- ( - 3 ) + 41}{2 \cdot 4}, k_{2} = \frac{- ( - 3 ) - 41}{2 \cdot 4}

k = \frac{- ( - 3 ) + 41}{2 \cdot 4} : \frac{3 + 41}{8}

k = \frac{- ( - 3 ) - 41}{2 \cdot 4} : \frac{3 - 41}{8}

The solutions to the quadratic equation are:

k = \frac{3 + 41}{8}, k = \frac{3 - 41}{8}

x^{2} = 10 - 3 x

9 x^{2} - 12 x + 13 = 0

3 x^{2} = 400

x, 4 x^{2} + y^{2} + 4 z^{2} - 4 y - 24 z + 36 = 0

4 x^{2} - 4 x + 1

What are the solutions to the equation -k^2-3k-2=-5k^2 ?
The solutions to the equation -k^2-3k-2=-5k^2 are k=(3+sqrt(41))/8 ,k=(3-sqrt(41))/8
Find the zeros of -k^2-3k-2=-5k^2
The zeros of -k^2-3k-2=-5k^2 are k=(3+sqrt(41))/8 ,k=(3-sqrt(41))/8