What are the solutions to the equation c^2+11c=12 ?

The solutions to the equation c^2+11c=12 are c=1,c=-12

Find the zeros of c^2+11c=12

The zeros of c^2+11c=12 are c=1,c=-12

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c^{2} + 11 c = 12

c = 1, c = - 12

Solution steps

c^{2} + 11 c = 12

Move

12

to the left side

c^{2} + 11 c - 12 = 0

Solve with the quadratic formula

c_{1, 2} = \frac{- 11 \pm 1 1 ^{2} - 4 \cdot 1 \cdot ( - 12 )}{2 \cdot 1}

1 1^{2} - 4 \cdot 1 \cdot (- 12) = 13

c_{1, 2} = \frac{- 11 \pm 13}{2 \cdot 1}

Separate the solutions

c_{1} = \frac{- 11 + 13}{2 \cdot 1}, c_{2} = \frac{- 11 - 13}{2 \cdot 1}

c = \frac{- 11 + 13}{2 \cdot 1} : 1

c = \frac{- 11 - 13}{2 \cdot 1} : - 12

The solutions to the quadratic equation are:

c = 1, c = - 12

\frac{1}{2} (4 k^{5})^{2}

- 3 + \frac{q}{2} < - 7

3 x^{3} + 3 x^{2} - 36 x

14 < - 1 - 3 y

\frac{1}{2} - \frac{2}{5}

What are the solutions to the equation c^2+11c=12 ?
The solutions to the equation c^2+11c=12 are c=1,c=-12
Find the zeros of c^2+11c=12
The zeros of c^2+11c=12 are c=1,c=-12