What are the solutions to the equation 12k^2+15k=16k+20 ?

The solutions to the equation 12k^2+15k=16k+20 are k= 4/3 ,k=-5/4

Find the zeros of 12k^2+15k=16k+20

The zeros of 12k^2+15k=16k+20 are k= 4/3 ,k=-5/4

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12 k^{2} + 15 k = 16 k + 20

k = \frac{4}{3}, k = - \frac{5}{4}

Decimal Notation

k = 1.33333 \dots, k = - 1.25

Solution steps

12 k^{2} + 15 k = 16 k + 20

Move

20

to the left side

12 k^{2} + 15 k - 20 = 16 k

Move

16 k

to the left side

12 k^{2} - k - 20 = 0

Solve with the quadratic formula

k_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 12 ( - 20 )}{2 \cdot 12}

(- 1)^{2} - 4 \cdot 12 (- 20) = 31

k_{1, 2} = \frac{- ( - 1 ) \pm 31}{2 \cdot 12}

Separate the solutions

k_{1} = \frac{- ( - 1 ) + 31}{2 \cdot 12}, k_{2} = \frac{- ( - 1 ) - 31}{2 \cdot 12}

k = \frac{- ( - 1 ) + 31}{2 \cdot 12} : \frac{4}{3}

k = \frac{- ( - 1 ) - 31}{2 \cdot 12} : - \frac{5}{4}

The solutions to the quadratic equation are:

k = \frac{4}{3}, k = - \frac{5}{4}

\frac{24 + 5 n - n ^{2}}{n ^{2} - 16 n + 64}

1 < ∣ x + 3 ∣ \leq 2

- x^{2} = - 9

(\frac{5 x ^{7}}{2})^{3}

2 x^{2} - 3 x + 4 = 0

What are the solutions to the equation 12k^2+15k=16k+20 ?
The solutions to the equation 12k^2+15k=16k+20 are k= 4/3 ,k=-5/4
Find the zeros of 12k^2+15k=16k+20
The zeros of 12k^2+15k=16k+20 are k= 4/3 ,k=-5/4