What are the solutions to the equation c(c-6)=27 ?

The solutions to the equation c(c-6)=27 are c=9,c=-3

Find the zeros of c(c-6)=27

The zeros of c(c-6)=27 are c=9,c=-3

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c (c - 6) = 27

c = 9, c = - 3

Solution steps

c (c - 6) = 27

Expand

c (c - 6) : c^{2} - 6 c

c^{2} - 6 c = 27

Move

27

to the left side

c^{2} - 6 c - 27 = 0

Solve with the quadratic formula

c_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 1 \cdot ( - 27 )}{2 \cdot 1}

(- 6)^{2} - 4 \cdot 1 \cdot (- 27) = 12

c_{1, 2} = \frac{- ( - 6 ) \pm 12}{2 \cdot 1}

Separate the solutions

c_{1} = \frac{- ( - 6 ) + 12}{2 \cdot 1}, c_{2} = \frac{- ( - 6 ) - 12}{2 \cdot 1}

c = \frac{- ( - 6 ) + 12}{2 \cdot 1} : 9

c = \frac{- ( - 6 ) - 12}{2 \cdot 1} : - 3

The solutions to the quadratic equation are:

c = 9, c = - 3

6 x + 2 = 2

x^{2} - 4 x = 5

\frac{2 π}{3} + 4 π

- 27 x - 30 x^{4}

(x^{- 4})^{5}

What are the solutions to the equation c(c-6)=27 ?
The solutions to the equation c(c-6)=27 are c=9,c=-3
Find the zeros of c(c-6)=27
The zeros of c(c-6)=27 are c=9,c=-3