What are the solutions to the equation 5k^2+6k-32=-12 ?

The solutions to the equation 5k^2+6k-32=-12 are k=(-3+sqrt(109))/5 ,k=-(3+sqrt(109))/5

Find the zeros of 5k^2+6k-32=-12

The zeros of 5k^2+6k-32=-12 are k=(-3+sqrt(109))/5 ,k=-(3+sqrt(109))/5

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5 k^{2} + 6 k - 32 = - 12

k = \frac{- 3 + 109}{5}, k = - \frac{3 + 109}{5}

Decimal Notation

k = 1.48806 \dots, k = - 2.68806 \dots

Solution steps

5 k^{2} + 6 k - 32 = - 12

Move

12

to the left side

5 k^{2} + 6 k - 20 = 0

Solve with the quadratic formula

k_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 5 ( - 20 )}{2 \cdot 5}

6^{2} - 4 \cdot 5 (- 20) = 2109

k_{1, 2} = \frac{- 6 \pm 2 109}{2 \cdot 5}

Separate the solutions

k_{1} = \frac{- 6 + 2 109}{2 \cdot 5}, k_{2} = \frac{- 6 - 2 109}{2 \cdot 5}

k = \frac{- 6 + 2 109}{2 \cdot 5} : \frac{- 3 + 109}{5}

k = \frac{- 6 - 2 109}{2 \cdot 5} : - \frac{3 + 109}{5}

The solutions to the quadratic equation are:

k = \frac{- 3 + 109}{5}, k = - \frac{3 + 109}{5}

(x + 2) (x - 3) - x (2 x + 1) + 6 x = 0

4 x^{2} - 2 = 94

5 x^{2} - 125 x = 0

p^{2} + 2 p - 8 = 0

x^{2} + 5 x + \frac{21}{4} = 0

What are the solutions to the equation 5k^2+6k-32=-12 ?
The solutions to the equation 5k^2+6k-32=-12 are k=(-3+sqrt(109))/5 ,k=-(3+sqrt(109))/5
Find the zeros of 5k^2+6k-32=-12
The zeros of 5k^2+6k-32=-12 are k=(-3+sqrt(109))/5 ,k=-(3+sqrt(109))/5