What are the solutions to the equation -4(x+3)^2=28 ?

The solutions to the equation -4(x+3)^2=28 are x=sqrt(7)i-3,x=-sqrt(7)i-3

Find the zeros of -4(x+3)^2=28

The zeros of -4(x+3)^2=28 are x=sqrt(7)i-3,x=-sqrt(7)i-3

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- 4 (x + 3)^{2} = 28

x = 7 i - 3, x = - 7 i - 3

Solution steps

- 4 (x + 3)^{2} = 28

Divide both sides by

- 4

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

For

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

the solutions are

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

Solve

x + 3 = - 7 : x = 7 i - 3

Solve

x + 3 = - - 7 : x = - 7 i - 3

The solutions to the quadratic equation are:

x = 7 i - 3, x = - 7 i - 3

6 (- 3 v + 1) = 5 (- 2 v - 2)

3 ∣ x - 1 ∣ - 8 > 16

3 x + 2 = \frac{2 x + 13}{3}

\frac{3}{5} (4 k - 1) \geq \frac{5}{3} (3 k - 2)

x^{3} + 7 x^{2} + 9 x + 63

What are the solutions to the equation -4(x+3)^2=28 ?
The solutions to the equation -4(x+3)^2=28 are x=sqrt(7)i-3,x=-sqrt(7)i-3
Find the zeros of -4(x+3)^2=28
The zeros of -4(x+3)^2=28 are x=sqrt(7)i-3,x=-sqrt(7)i-3