What are the solutions to the equation (w+4)^2+95=5 ?

The solutions to the equation (w+4)^2+95=5 are w=3sqrt(10)i-4,w=-3sqrt(10)i-4

Find the zeros of (w+4)^2+95=5

The zeros of (w+4)^2+95=5 are w=3sqrt(10)i-4,w=-3sqrt(10)i-4

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(w + 4)^{2} + 95 = 5

w = 310 i - 4, w = - 310 i - 4

Solution steps

(w + 4)^{2} + 95 = 5

Move

95

to the right side

(w + 4)^{2} = - 90

For

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

the solutions are

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

Solve

w + 4 = - 90 : w = 310 i - 4

Solve

w + 4 = - - 90 : w = - 310 i - 4

The solutions to the quadratic equation are:

w = 310 i - 4, w = - 310 i - 4

∣ 0.5 x + 2.5 ∣ > 1.5

\frac{2 + i}{2 - i}

2 x + 9 = 14

- 5 (- 2 x - 5) - x - 1 = - 12

28 + 12 x = 8 x - 4

What are the solutions to the equation (w+4)^2+95=5 ?
The solutions to the equation (w+4)^2+95=5 are w=3sqrt(10)i-4,w=-3sqrt(10)i-4
Find the zeros of (w+4)^2+95=5
The zeros of (w+4)^2+95=5 are w=3sqrt(10)i-4,w=-3sqrt(10)i-4