What are the solutions to the equation 2/5 (x+3)^2+8=-32 ?

The solutions to the equation 2/5 (x+3)^2+8=-32 are x=10i-3,x=-10i-3

Find the zeros of 2/5 (x+3)^2+8=-32

The zeros of 2/5 (x+3)^2+8=-32 are x=10i-3,x=-10i-3

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\frac{2}{5} (x + 3)^{2} + 8 = - 32

x = 10 i - 3, x = - 10 i - 3

Solution steps

\frac{2}{5} (x + 3)^{2} + 8 = - 32

Move

8

to the right side

\frac{2}{5} (x + 3)^{2} = - 40

Multiply both sides by

5

2 (x + 3)^{2} = - 200

Divide both sides by

2

(x + 3)^{2} = - 100

For

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

the solutions are

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

Solve

x + 3 = - 100 : x = 10 i - 3

Solve

x + 3 = - - 100 : x = - 10 i - 3

The solutions to the quadratic equation are:

x = 10 i - 3, x = - 10 i - 3

0^{2} + y^{2} = 49

x^{2} + 18 x - 16 = 0

\frac{1}{2} - 2

- 120 \cdot \frac{π}{180}

- 6 (u + 1) = 2 (1 - 3 u) - 8

What are the solutions to the equation 2/5 (x+3)^2+8=-32 ?
The solutions to the equation 2/5 (x+3)^2+8=-32 are x=10i-3,x=-10i-3
Find the zeros of 2/5 (x+3)^2+8=-32
The zeros of 2/5 (x+3)^2+8=-32 are x=10i-3,x=-10i-3