What are the solutions to the equation 5(3x-2)^2+27=32 ?

The solutions to the equation 5(3x-2)^2+27=32 are x=1,x= 1/3

Find the zeros of 5(3x-2)^2+27=32

The zeros of 5(3x-2)^2+27=32 are x=1,x= 1/3

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5 (3 x - 2)^{2} + 27 = 32

x = 1, x = \frac{1}{3}

Decimal Notation

x = 1, x = 0.33333 \dots

Solution steps

5 (3 x - 2)^{2} + 27 = 32

Move

27

to the right side

5 (3 x - 2)^{2} = 5

Divide both sides by

5

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

For

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

the solutions are

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

Solve

3 x - 2 = 1 : x = 1

Solve

3 x - 2 = - 1 : x = \frac{1}{3}

The solutions to the quadratic equation are:

x = 1, x = \frac{1}{3}

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

4 y^{2} = - 12 y - 9

x^{2} = - 9 x - 20

19.25 = - 8.5 (2 r - 1.75)^{2}

x, x (2 x + 1) = 0

What are the solutions to the equation 5(3x-2)^2+27=32 ?
The solutions to the equation 5(3x-2)^2+27=32 are x=1,x= 1/3
Find the zeros of 5(3x-2)^2+27=32
The zeros of 5(3x-2)^2+27=32 are x=1,x= 1/3