What are the solutions to the equation 21-8k-2k^2=2k^2 ?

The solutions to the equation 21-8k-2k^2=2k^2 are k=-7/2 ,k= 3/2

Find the zeros of 21-8k-2k^2=2k^2

The zeros of 21-8k-2k^2=2k^2 are k=-7/2 ,k= 3/2

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21 - 8 k - 2 k^{2} = 2 k^{2}

k = - \frac{7}{2}, k = \frac{3}{2}

Decimal Notation

k = - 3.5, k = 1.5

Solution steps

21 - 8 k - 2 k^{2} = 2 k^{2}

Move

2 k^{2}

to the left side

21 - 8 k - 4 k^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

- 4 k^{2} - 8 k + 21 = 0

Solve with the quadratic formula

k_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 ( - 4 ) \cdot 21}{2 ( - 4 )}

(- 8)^{2} - 4 (- 4) \cdot 21 = 20

k_{1, 2} = \frac{- ( - 8 ) \pm 20}{2 ( - 4 )}

Separate the solutions

k_{1} = \frac{- ( - 8 ) + 20}{2 ( - 4 )}, k_{2} = \frac{- ( - 8 ) - 20}{2 ( - 4 )}

k = \frac{- ( - 8 ) + 20}{2 ( - 4 )} : - \frac{7}{2}

k = \frac{- ( - 8 ) - 20}{2 ( - 4 )} : \frac{3}{2}

The solutions to the quadratic equation are:

k = - \frac{7}{2}, k = \frac{3}{2}

x^{2} + 12 x + 20 = 0

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

y, 9 x = y^{2} + 18

(1 - (- 1)) x^{2} - (- 1 + 1) x - 2 = 0

x^{2} + 16 x - 1 = 0

What are the solutions to the equation 21-8k-2k^2=2k^2 ?
The solutions to the equation 21-8k-2k^2=2k^2 are k=-7/2 ,k= 3/2
Find the zeros of 21-8k-2k^2=2k^2
The zeros of 21-8k-2k^2=2k^2 are k=-7/2 ,k= 3/2