What are the solutions to the equation 3(2b+3)^2=36 ?

The solutions to the equation 3(2b+3)^2=36 are b=(2sqrt(3)-3)/2 ,b=(-2sqrt(3)-3)/2

Find the zeros of 3(2b+3)^2=36

The zeros of 3(2b+3)^2=36 are b=(2sqrt(3)-3)/2 ,b=(-2sqrt(3)-3)/2

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3 (2 b + 3)^{2} = 36

b = \frac{2 3 - 3}{2}, b = \frac{- 2 3 - 3}{2}

Decimal Notation

b = 0.23205 \dots, b = - 3.23205 \dots

Solution steps

3 (2 b + 3)^{2} = 36

Divide both sides by

3

(2 b + 3)^{2} = 12

For

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

the solutions are

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

Solve

2 b + 3 = 12 : b = \frac{2 3 - 3}{2}

Solve

2 b + 3 = - 12 : b = \frac{- 2 3 - 3}{2}

The solutions to the quadratic equation are:

b = \frac{2 3 - 3}{2}, b = \frac{- 2 3 - 3}{2}

(x - a)^{2} = b^{2}

- 3 y^{2} + 5 y - 15 = - 4 y^{2}

(- x + 6)^{2} - 12 = 0

- x^{2} - 70 x + 2139 = 0

25 y^{2} - 9 y - 180 = 0

What are the solutions to the equation 3(2b+3)^2=36 ?
The solutions to the equation 3(2b+3)^2=36 are b=(2sqrt(3)-3)/2 ,b=(-2sqrt(3)-3)/2
Find the zeros of 3(2b+3)^2=36
The zeros of 3(2b+3)^2=36 are b=(2sqrt(3)-3)/2 ,b=(-2sqrt(3)-3)/2