What are the solutions to the equation 29=3(x+7)^2+41 ?

The solutions to the equation 29=3(x+7)^2+41 are x=2i-7,x=-2i-7

Find the zeros of 29=3(x+7)^2+41

The zeros of 29=3(x+7)^2+41 are x=2i-7,x=-2i-7

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29 = 3 (x + 7)^{2} + 41

x = 2 i - 7, x = - 2 i - 7

Solution steps

29 = 3 (x + 7)^{2} + 41

Switch sides

3 (x + 7)^{2} + 41 = 29

Move

41

to the right side

3 (x + 7)^{2} = - 12

Divide both sides by

3

(x + 7)^{2} = - 4

For

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

the solutions are

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

Solve

x + 7 = - 4 : x = 2 i - 7

Solve

x + 7 = - - 4 : x = - 2 i - 7

The solutions to the quadratic equation are:

x = 2 i - 7, x = - 2 i - 7

3^{2} + y^{2} = 25

(x + 3)^{2} - (x - 1)^{2} = x^{2}

3 x^{2} + 16 x - 3 = 0

8 x^{2} + 9 x + 1 = 0

2 x^{2} - x - 6 = 0

What are the solutions to the equation 29=3(x+7)^2+41 ?
The solutions to the equation 29=3(x+7)^2+41 are x=2i-7,x=-2i-7
Find the zeros of 29=3(x+7)^2+41
The zeros of 29=3(x+7)^2+41 are x=2i-7,x=-2i-7