What are the solutions to the equation c^2+7c+12=0 ?

The solutions to the equation c^2+7c+12=0 are c=-3,c=-4

Find the zeros of c^2+7c+12=0

The zeros of c^2+7c+12=0 are c=-3,c=-4

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c^{2} + 7 c + 12 = 0

c = - 3, c = - 4

Solution steps

c^{2} + 7 c + 12 = 0

Solve with the quadratic formula

c_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 1 \cdot 12}{2 \cdot 1}

7^{2} - 4 \cdot 1 \cdot 12 = 1

c_{1, 2} = \frac{- 7 \pm 1}{2 \cdot 1}

Separate the solutions

c_{1} = \frac{- 7 + 1}{2 \cdot 1}, c_{2} = \frac{- 7 - 1}{2 \cdot 1}

c = \frac{- 7 + 1}{2 \cdot 1} : - 3

c = \frac{- 7 - 1}{2 \cdot 1} : - 4

The solutions to the quadratic equation are:

c = - 3, c = - 4

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What are the solutions to the equation c^2+7c+12=0 ?
The solutions to the equation c^2+7c+12=0 are c=-3,c=-4
Find the zeros of c^2+7c+12=0
The zeros of c^2+7c+12=0 are c=-3,c=-4