What are the solutions to the equation 4(x+5)^2+20=12 ?

The solutions to the equation 4(x+5)^2+20=12 are x=sqrt(2)i-5,x=-sqrt(2)i-5

Find the zeros of 4(x+5)^2+20=12

The zeros of 4(x+5)^2+20=12 are x=sqrt(2)i-5,x=-sqrt(2)i-5

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4 (x + 5)^{2} + 20 = 12

x = 2 i - 5, x = - 2 i - 5

Solution steps

4 (x + 5)^{2} + 20 = 12

Move

20

to the right side

4 (x + 5)^{2} = - 8

Divide both sides by

4

(x + 5)^{2} = - 2

For

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

the solutions are

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

Solve

x + 5 = - 2 : x = 2 i - 5

Solve

x + 5 = - - 2 : x = - 2 i - 5

The solutions to the quadratic equation are:

x = 2 i - 5, x = - 2 i - 5

2 x (x - 2) = 10 x^{2} - 5 x

0.1 x^{2} - 70 x + 12000 = 0

7 k^{2} + 12 = 25 k

(x - 1)^{2} - 2 (x - 1) = 15

(2 x - 3) (7 x + 4) = 0

What are the solutions to the equation 4(x+5)^2+20=12 ?
The solutions to the equation 4(x+5)^2+20=12 are x=sqrt(2)i-5,x=-sqrt(2)i-5
Find the zeros of 4(x+5)^2+20=12
The zeros of 4(x+5)^2+20=12 are x=sqrt(2)i-5,x=-sqrt(2)i-5