What are the solutions to the equation 5c(c+1)+6=0 ?

The solutions to the equation 5c(c+1)+6=0 are c=-1/2+i(sqrt(95))/(10),c=-1/2-i(sqrt(95))/(10)

Find the zeros of 5c(c+1)+6=0

The zeros of 5c(c+1)+6=0 are c=-1/2+i(sqrt(95))/(10),c=-1/2-i(sqrt(95))/(10)

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5 c (c + 1) + 6 = 0

c = - \frac{1}{2} + i \frac{95}{10}, c = - \frac{1}{2} - i \frac{95}{10}

Solution steps

5 c (c + 1) + 6 = 0

Expand

5 c (c + 1) + 6 : 5 c^{2} + 5 c + 6

5 c^{2} + 5 c + 6 = 0

Solve with the quadratic formula

c_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 5 \cdot 6}{2 \cdot 5}

Simplify

5^{2} - 4 \cdot 5 \cdot 6 : 95 i

c_{1, 2} = \frac{- 5 \pm 95 i}{2 \cdot 5}

Separate the solutions

c_{1} = \frac{- 5 + 95 i}{2 \cdot 5}, c_{2} = \frac{- 5 - 95 i}{2 \cdot 5}

c = \frac{- 5 + 95 i}{2 \cdot 5} : - \frac{1}{2} + i \frac{95}{10}

c = \frac{- 5 - 95 i}{2 \cdot 5} : - \frac{1}{2} - i \frac{95}{10}

The solutions to the quadratic equation are:

c = - \frac{1}{2} + i \frac{95}{10}, c = - \frac{1}{2} - i \frac{95}{10}

4 \cdot 4

k + 4 - 2 (k - 12) > 0

- 3 x^{2} - 18 x - 15

7 x + 7 - 5 x = 3

\frac{( \frac{π}{2} )}{2}

What are the solutions to the equation 5c(c+1)+6=0 ?
The solutions to the equation 5c(c+1)+6=0 are c=-1/2+i(sqrt(95))/(10),c=-1/2-i(sqrt(95))/(10)
Find the zeros of 5c(c+1)+6=0
The zeros of 5c(c+1)+6=0 are c=-1/2+i(sqrt(95))/(10),c=-1/2-i(sqrt(95))/(10)