What are the solutions to the equation 6(x+3)^2+20=12 ?

The solutions to the equation 6(x+3)^2+20=12 are x=-3+i(2sqrt(3))/3 ,x=-3-i(2sqrt(3))/3

Find the zeros of 6(x+3)^2+20=12

The zeros of 6(x+3)^2+20=12 are x=-3+i(2sqrt(3))/3 ,x=-3-i(2sqrt(3))/3

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6 (x + 3)^{2} + 20 = 12

x = - 3 + i \frac{2 3}{3}, x = - 3 - i \frac{2 3}{3}

Solution steps

6 (x + 3)^{2} + 20 = 12

Move

20

to the right side

6 (x + 3)^{2} = - 8

Divide both sides by

6

(x + 3)^{2} = - \frac{4}{3}

For

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

the solutions are

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

Solve

x + 3 = - \frac{4}{3} : x = - 3 + i \frac{2 3}{3}

Solve

x + 3 = - - \frac{4}{3} : x = - 3 - i \frac{2 3}{3}

The solutions to the quadratic equation are:

x = - 3 + i \frac{2 3}{3}, x = - 3 - i \frac{2 3}{3}

2 (x - 5)^{2} + 50 = 0

x^{2} + 2 x - 1 = 0

5 x^{2} - 6 = 29

(4 + y) (5 y + 8) = 0

12 + 2.4 x - 3 x^{2} = 0

What are the solutions to the equation 6(x+3)^2+20=12 ?
The solutions to the equation 6(x+3)^2+20=12 are x=-3+i(2sqrt(3))/3 ,x=-3-i(2sqrt(3))/3
Find the zeros of 6(x+3)^2+20=12
The zeros of 6(x+3)^2+20=12 are x=-3+i(2sqrt(3))/3 ,x=-3-i(2sqrt(3))/3