What are the solutions to the equation (5c+1)^2=-27 ?

The solutions to the equation (5c+1)^2=-27 are c=-1/5+i(3sqrt(3))/5 ,c=-1/5-i(3sqrt(3))/5

Find the zeros of (5c+1)^2=-27

The zeros of (5c+1)^2=-27 are c=-1/5+i(3sqrt(3))/5 ,c=-1/5-i(3sqrt(3))/5

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(5 c + 1)^{2} = - 27

c = - \frac{1}{5} + i \frac{3 3}{5}, c = - \frac{1}{5} - i \frac{3 3}{5}

Solution steps

(5 c + 1)^{2} = - 27

For

(g (x))^{2} = f (a)

the solutions are

g (x) = f (a), - f (a)

Solve

5 c + 1 = - 27 : c = - \frac{1}{5} + i \frac{3 3}{5}

Solve

5 c + 1 = - - 27 : c = - \frac{1}{5} - i \frac{3 3}{5}

The solutions to the quadratic equation are:

c = - \frac{1}{5} + i \frac{3 3}{5}, c = - \frac{1}{5} - i \frac{3 3}{5}

3 x^{2} + 6 x + 18 = 0

a^{2} + 4 a + 1 = 0

4 x^{2} - 6 x + 16 = 0

x^{2} - 14 x + 67 = 0

2 x^{2} - 4 x = 0

What are the solutions to the equation (5c+1)^2=-27 ?
The solutions to the equation (5c+1)^2=-27 are c=-1/5+i(3sqrt(3))/5 ,c=-1/5-i(3sqrt(3))/5
Find the zeros of (5c+1)^2=-27
The zeros of (5c+1)^2=-27 are c=-1/5+i(3sqrt(3))/5 ,c=-1/5-i(3sqrt(3))/5