What are the solutions to the equation 3(k^2-36)+k^2=6k ?

The solutions to the equation 3(k^2-36)+k^2=6k are k=6,k=-9/2

Find the zeros of 3(k^2-36)+k^2=6k

The zeros of 3(k^2-36)+k^2=6k are k=6,k=-9/2

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3 (k^{2} - 36) + k^{2} = 6 k

k = 6, k = - \frac{9}{2}

Decimal Notation

k = 6, k = - 4.5

Solution steps

3 (k^{2} - 36) + k^{2} = 6 k

Expand

3 (k^{2} - 36) + k^{2} : 4 k^{2} - 108

4 k^{2} - 108 = 6 k

Move

6 k

to the left side

4 k^{2} - 108 - 6 k = 0

Write in the standard form

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

4 k^{2} - 6 k - 108 = 0

Solve with the quadratic formula

k_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 4 ( - 108 )}{2 \cdot 4}

(- 6)^{2} - 4 \cdot 4 (- 108) = 42

k_{1, 2} = \frac{- ( - 6 ) \pm 42}{2 \cdot 4}

Separate the solutions

k_{1} = \frac{- ( - 6 ) + 42}{2 \cdot 4}, k_{2} = \frac{- ( - 6 ) - 42}{2 \cdot 4}

k = \frac{- ( - 6 ) + 42}{2 \cdot 4} : 6

k = \frac{- ( - 6 ) - 42}{2 \cdot 4} : - \frac{9}{2}

The solutions to the quadratic equation are:

k = 6, k = - \frac{9}{2}

x, f = x^{2} y^{3} + 3 y - 5 x

x^{2} - 5 = - 54

(x + 6) (x + 6) = 13

(x + 2)^{2} + 25 = 0

2 x^{2} + 1 = 17

What are the solutions to the equation 3(k^2-36)+k^2=6k ?
The solutions to the equation 3(k^2-36)+k^2=6k are k=6,k=-9/2
Find the zeros of 3(k^2-36)+k^2=6k
The zeros of 3(k^2-36)+k^2=6k are k=6,k=-9/2