What are the solutions to the equation 3k(k-4)+k+4(k+1)=0 ?

The solutions to the equation 3k(k-4)+k+4(k+1)=0 are k= 4/3 ,k=1

Find the zeros of 3k(k-4)+k+4(k+1)=0

The zeros of 3k(k-4)+k+4(k+1)=0 are k= 4/3 ,k=1

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3 k (k - 4) + k + 4 (k + 1) = 0

k = \frac{4}{3}, k = 1

Decimal Notation

k = 1.33333 \dots, k = 1

Solution steps

3 k (k - 4) + k + 4 (k + 1) = 0

Expand

3 k (k - 4) + k + 4 (k + 1) : 3 k^{2} - 7 k + 4

3 k^{2} - 7 k + 4 = 0

Solve with the quadratic formula

k_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 3 \cdot 4}{2 \cdot 3}

(- 7)^{2} - 4 \cdot 3 \cdot 4 = 1

k_{1, 2} = \frac{- ( - 7 ) \pm 1}{2 \cdot 3}

Separate the solutions

k_{1} = \frac{- ( - 7 ) + 1}{2 \cdot 3}, k_{2} = \frac{- ( - 7 ) - 1}{2 \cdot 3}

k = \frac{- ( - 7 ) + 1}{2 \cdot 3} : \frac{4}{3}

k = \frac{- ( - 7 ) - 1}{2 \cdot 3} : 1

The solutions to the quadratic equation are:

k = \frac{4}{3}, k = 1

(2 x + 3) (x - 5) = 0

m^{2} - 6 m + 5 = 0

x^{2} - 2 x - 960 = 0

0 = x^{2} + 5 x + 4

- x^{2} + 2 x

What are the solutions to the equation 3k(k-4)+k+4(k+1)=0 ?
The solutions to the equation 3k(k-4)+k+4(k+1)=0 are k= 4/3 ,k=1
Find the zeros of 3k(k-4)+k+4(k+1)=0
The zeros of 3k(k-4)+k+4(k+1)=0 are k= 4/3 ,k=1