What are the solutions to the equation 15=c*(c+2) ?

The solutions to the equation 15=c*(c+2) are c=3,c=-5

Find the zeros of 15=c*(c+2)

The zeros of 15=c*(c+2) are c=3,c=-5

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15 = c \cdot (c + 2)

c = 3, c = - 5

Solution steps

15 = c (c + 2)

Expand

c (c + 2) : c^{2} + 2 c

15 = c^{2} + 2 c

Switch sides

c^{2} + 2 c = 15

Move

15

to the left side

c^{2} + 2 c - 15 = 0

Solve with the quadratic formula

c_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot ( - 15 )}{2 \cdot 1}

2^{2} - 4 \cdot 1 \cdot (- 15) = 8

c_{1, 2} = \frac{- 2 \pm 8}{2 \cdot 1}

Separate the solutions

c_{1} = \frac{- 2 + 8}{2 \cdot 1}, c_{2} = \frac{- 2 - 8}{2 \cdot 1}

c = \frac{- 2 + 8}{2 \cdot 1} : 3

c = \frac{- 2 - 8}{2 \cdot 1} : - 5

The solutions to the quadratic equation are:

c = 3, c = - 5

3 x^{2} - 16 x = 0

\frac{2}{x + 3} > \frac{1}{x - 2}

1 - \frac{π}{2}

∣ z ∣ \leq 26

9^{2} + b^{2} = 1 4^{2}

What are the solutions to the equation 15=c*(c+2) ?
The solutions to the equation 15=c*(c+2) are c=3,c=-5
Find the zeros of 15=c*(c+2)
The zeros of 15=c*(c+2) are c=3,c=-5