What are the solutions to the equation c^2+2c=48 ?

The solutions to the equation c^2+2c=48 are c=6,c=-8

Find the zeros of c^2+2c=48

The zeros of c^2+2c=48 are c=6,c=-8

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c^{2} + 2 c = 48

c = 6, c = - 8

Solution steps

c^{2} + 2 c = 48

Move

48

to the left side

c^{2} + 2 c - 48 = 0

Solve with the quadratic formula

c_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot ( - 48 )}{2 \cdot 1}

2^{2} - 4 \cdot 1 \cdot (- 48) = 14

c_{1, 2} = \frac{- 2 \pm 14}{2 \cdot 1}

Separate the solutions

c_{1} = \frac{- 2 + 14}{2 \cdot 1}, c_{2} = \frac{- 2 - 14}{2 \cdot 1}

c = \frac{- 2 + 14}{2 \cdot 1} : 6

c = \frac{- 2 - 14}{2 \cdot 1} : - 8

The solutions to the quadratic equation are:

c = 6, c = - 8

6 r + r^{2} = 27

4 x^{2} - 16 x - 20 = 0

3 q^{2} - 110 q + 900 = 0

(x - 2) (x + 3) = 2 x^{2} - 4 x

9 x^{2} - 9 x + 19 = 0

What are the solutions to the equation c^2+2c=48 ?
The solutions to the equation c^2+2c=48 are c=6,c=-8
Find the zeros of c^2+2c=48
The zeros of c^2+2c=48 are c=6,c=-8