What are the solutions to the equation (v+1/5)^2-9=0 ?

The solutions to the equation (v+1/5)^2-9=0 are v= 14/5 ,v=-16/5

Find the zeros of (v+1/5)^2-9=0

The zeros of (v+1/5)^2-9=0 are v= 14/5 ,v=-16/5

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(v + \frac{1}{5})^{2} - 9 = 0

v = \frac{14}{5}, v = - \frac{16}{5}

Decimal Notation

v = 2.8, v = - 3.2

Solution steps

(v + \frac{1}{5})^{2} - 9 = 0

Move

9

to the right side

(v + \frac{1}{5})^{2} = 9

For

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

the solutions are

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

Solve

v + \frac{1}{5} = 9 : v = \frac{14}{5}

Solve

v + \frac{1}{5} = - 9 : v = - \frac{16}{5}

The solutions to the quadratic equation are:

v = \frac{14}{5}, v = - \frac{16}{5}

5 x^{2} - 8 x + 2 = 0

\frac{3}{2} p - 5 + \frac{9}{4} p = 7 - \frac{5}{4} p

10 x^{2} + 13 x + 3

4 x + 1 = x - 8

0.9 x - 5.9 = 0.4

What are the solutions to the equation (v+1/5)^2-9=0 ?
The solutions to the equation (v+1/5)^2-9=0 are v= 14/5 ,v=-16/5
Find the zeros of (v+1/5)^2-9=0
The zeros of (v+1/5)^2-9=0 are v= 14/5 ,v=-16/5