What are the solutions to the equation c^2+12c+36=121 ?

The solutions to the equation c^2+12c+36=121 are c=5,c=-17

Find the zeros of c^2+12c+36=121

The zeros of c^2+12c+36=121 are c=5,c=-17

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c^{2} + 12 c + 36 = 121

c = 5, c = - 17

Solution steps

c^{2} + 12 c + 36 = 121

Move

121

to the left side

c^{2} + 12 c - 85 = 0

Solve with the quadratic formula

c_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 1 \cdot ( - 85 )}{2 \cdot 1}

1 2^{2} - 4 \cdot 1 \cdot (- 85) = 22

c_{1, 2} = \frac{- 12 \pm 22}{2 \cdot 1}

Separate the solutions

c_{1} = \frac{- 12 + 22}{2 \cdot 1}, c_{2} = \frac{- 12 - 22}{2 \cdot 1}

c = \frac{- 12 + 22}{2 \cdot 1} : 5

c = \frac{- 12 - 22}{2 \cdot 1} : - 17

The solutions to the quadratic equation are:

c = 5, c = - 17

r^{2} - 16 r + 35 = 7

1 3^{2} = x^{2} + (x + 7)^{2}

x^{2} - 11 x - 12

(x + 3) (x + 1) = 0

4 x^{2} + 12 x + 36 = 0

What are the solutions to the equation c^2+12c+36=121 ?
The solutions to the equation c^2+12c+36=121 are c=5,c=-17
Find the zeros of c^2+12c+36=121
The zeros of c^2+12c+36=121 are c=5,c=-17