What are the solutions to the equation 4(x+2)^2+320=0 ?

The solutions to the equation 4(x+2)^2+320=0 are x=4sqrt(5)i-2,x=-4sqrt(5)i-2

Find the zeros of 4(x+2)^2+320=0

The zeros of 4(x+2)^2+320=0 are x=4sqrt(5)i-2,x=-4sqrt(5)i-2

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4 (x + 2)^{2} + 320 = 0

x = 45 i - 2, x = - 45 i - 2

Solution steps

4 (x + 2)^{2} + 320 = 0

Move

320

to the right side

4 (x + 2)^{2} = - 320

Divide both sides by

4

(x + 2)^{2} = - 80

For

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

the solutions are

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

Solve

x + 2 = - 80 : x = 45 i - 2

Solve

x + 2 = - - 80 : x = - 45 i - 2

The solutions to the quadratic equation are:

x = 45 i - 2, x = - 45 i - 2

3 (8 x - 6) + 9 = 63

- 9 (\frac{1}{2})

\frac{[ 4 ( 1 + h ) - ( 1 + h ) ^{2} ] - 3}{h}

1536

- 2 \frac{x + 1}{3} = 2 - x

What are the solutions to the equation 4(x+2)^2+320=0 ?
The solutions to the equation 4(x+2)^2+320=0 are x=4sqrt(5)i-2,x=-4sqrt(5)i-2
Find the zeros of 4(x+2)^2+320=0
The zeros of 4(x+2)^2+320=0 are x=4sqrt(5)i-2,x=-4sqrt(5)i-2