What are the solutions to the equation 9(z+1)^2+12=16 ?

The solutions to the equation 9(z+1)^2+12=16 are z=-1/3 ,z=-5/3

Find the zeros of 9(z+1)^2+12=16

The zeros of 9(z+1)^2+12=16 are z=-1/3 ,z=-5/3

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9 (z + 1)^{2} + 12 = 16

z = - \frac{1}{3}, z = - \frac{5}{3}

Solution steps

9 (z + 1)^{2} + 12 = 16

Move

12

to the right side

9 (z + 1)^{2} = 4

Divide both sides by

9

(z + 1)^{2} = \frac{4}{9}

For

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

the solutions are

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

Solve

z + 1 = \frac{4}{9} : z = - \frac{1}{3}

Solve

z + 1 = - \frac{4}{9} : z = - \frac{5}{3}

The solutions to the quadratic equation are:

z = - \frac{1}{3}, z = - \frac{5}{3}

2 (3 x - 7) = 16

8 \div 2 (1 + 3)

0 = 4 x^{2} + 3 x + 8

- 2 x > - 3

62 5^{\frac{1}{2}}

What are the solutions to the equation 9(z+1)^2+12=16 ?
The solutions to the equation 9(z+1)^2+12=16 are z=-1/3 ,z=-5/3
Find the zeros of 9(z+1)^2+12=16
The zeros of 9(z+1)^2+12=16 are z=-1/3 ,z=-5/3