What are the solutions to the equation -2(p+4)^2=24 ?

The solutions to the equation -2(p+4)^2=24 are p=2sqrt(3)i-4,p=-2sqrt(3)i-4

Find the zeros of -2(p+4)^2=24

The zeros of -2(p+4)^2=24 are p=2sqrt(3)i-4,p=-2sqrt(3)i-4

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- 2 (p + 4)^{2} = 24

p = 23 i - 4, p = - 23 i - 4

Solution steps

- 2 (p + 4)^{2} = 24

Divide both sides by

- 2

(p + 4)^{2} = - 12

For

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

the solutions are

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

Solve

p + 4 = - 12 : p = 23 i - 4

Solve

p + 4 = - - 12 : p = - 23 i - 4

The solutions to the quadratic equation are:

p = 23 i - 4, p = - 23 i - 4

lo g_{b} (\frac{b}{x})

r^{2} + 9 r + 18 = 0

2^{\frac{1}{2}} \cdot 2^{\frac{1}{4}}

∣ 5 x + 1 ∣ = 36

- 28 - 7 x \leq 0

What are the solutions to the equation -2(p+4)^2=24 ?
The solutions to the equation -2(p+4)^2=24 are p=2sqrt(3)i-4,p=-2sqrt(3)i-4
Find the zeros of -2(p+4)^2=24
The zeros of -2(p+4)^2=24 are p=2sqrt(3)i-4,p=-2sqrt(3)i-4