What are the solutions to the equation (x+b/3)^2-2-(b^2)/9 =0 ?

The solutions to the equation (x+b/3)^2-2-(b^2)/9 =0 are x=(sqrt(b^2+18)-b)/3 ,x=(-sqrt(b^2+18)-b)/3

Find the zeros of (x+b/3)^2-2-(b^2)/9 =0

The zeros of (x+b/3)^2-2-(b^2)/9 =0 are x=(sqrt(b^2+18)-b)/3 ,x=(-sqrt(b^2+18)-b)/3

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(x + \frac{b}{3})^{2} - 2 - \frac{b ^{2}}{9} = 0

x = \frac{b ^{2} + 18 - b}{3}, x = \frac{- b ^{2} + 18 - b}{3}

Solution steps

(x + \frac{b}{3})^{2} - 2 - \frac{b ^{2}}{9} = 0

Move

\frac{b ^{2}}{9}

to the right side

(x + \frac{b}{3})^{2} - 2 = \frac{b ^{2}}{9}

Move

2

to the right side

(x + \frac{b}{3})^{2} = \frac{b ^{2}}{9} + 2

For

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

the solutions are

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

Solve

x + \frac{b}{3} = \frac{b ^{2}}{9} + 2 : x = \frac{b ^{2} + 18 - b}{3}

Solve

x + \frac{b}{3} = - \frac{b ^{2}}{9} + 2 : x = \frac{- b ^{2} + 18 - b}{3}

The solutions to the quadratic equation are:

x = \frac{b ^{2} + 18 - b}{3}, x = \frac{- b ^{2} + 18 - b}{3}

2 x^{2} - x - 2

(2 x + 3)^{2} = 4

x^{2} + 4 x + 4 = 2 x + 7

t, f = k t^{2} + 2

25 x^{2} + 30 x = - 5

What are the solutions to the equation (x+b/3)^2-2-(b^2)/9 =0 ?
The solutions to the equation (x+b/3)^2-2-(b^2)/9 =0 are x=(sqrt(b^2+18)-b)/3 ,x=(-sqrt(b^2+18)-b)/3
Find the zeros of (x+b/3)^2-2-(b^2)/9 =0
The zeros of (x+b/3)^2-2-(b^2)/9 =0 are x=(sqrt(b^2+18)-b)/3 ,x=(-sqrt(b^2+18)-b)/3