What are the solutions to the equation c^2+4c-17=-6 ?

The solutions to the equation c^2+4c-17=-6 are c=-2+sqrt(15),c=-2-sqrt(15)

Find the zeros of c^2+4c-17=-6

The zeros of c^2+4c-17=-6 are c=-2+sqrt(15),c=-2-sqrt(15)

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c^{2} + 4 c - 17 = - 6

c = - 2 + 15, c = - 2 - 15

Solution steps

c^{2} + 4 c - 17 = - 6

Move

6

to the left side

c^{2} + 4 c - 11 = 0

Solve with the quadratic formula

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

4^{2} - 4 \cdot 1 \cdot (- 11) = 215

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

Separate the solutions

c_{1} = \frac{- 4 + 2 15}{2 \cdot 1}, c_{2} = \frac{- 4 - 2 15}{2 \cdot 1}

c = \frac{- 4 + 2 15}{2 \cdot 1} : - 2 + 15

c = \frac{- 4 - 2 15}{2 \cdot 1} : - 2 - 15

The solutions to the quadratic equation are:

c = - 2 + 15, c = - 2 - 15

x > 3 or x < 5

x^{3} + 5 x^{2} + 4 x + 20

(\frac{6}{10})^{2}

(\frac{x}{3})^{3} (9 x)^{- 2}

(\frac{8}{3})^{2}

What are the solutions to the equation c^2+4c-17=-6 ?
The solutions to the equation c^2+4c-17=-6 are c=-2+sqrt(15),c=-2-sqrt(15)
Find the zeros of c^2+4c-17=-6
The zeros of c^2+4c-17=-6 are c=-2+sqrt(15),c=-2-sqrt(15)