What are the solutions to the equation 3(3x+4)^2+18=27 ?

The solutions to the equation 3(3x+4)^2+18=27 are x=(sqrt(3)-4)/3 ,x=(-sqrt(3)-4)/3

Find the zeros of 3(3x+4)^2+18=27

The zeros of 3(3x+4)^2+18=27 are x=(sqrt(3)-4)/3 ,x=(-sqrt(3)-4)/3

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3 (3 x + 4)^{2} + 18 = 27

x = \frac{3 - 4}{3}, x = \frac{- 3 - 4}{3}

Decimal Notation

x = - 0.75598 \dots, x = - 1.91068 \dots

Solution steps

3 (3 x + 4)^{2} + 18 = 27

Move

18

to the right side

3 (3 x + 4)^{2} = 9

Divide both sides by

3

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

For

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

the solutions are

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

Solve

3 x + 4 = 3 : x = \frac{3 - 4}{3}

Solve

3 x + 4 = - 3 : x = \frac{- 3 - 4}{3}

The solutions to the quadratic equation are:

x = \frac{3 - 4}{3}, x = \frac{- 3 - 4}{3}

u^{2} - 5 u + 6 = 0

x^{2} - 8 x - 17 = 0

- 20 = 8 x - 4 x^{2} - 3

x^{2} + 6 x = - 15

r^{2} + 2 r - 20 = - 3

What are the solutions to the equation 3(3x+4)^2+18=27 ?
The solutions to the equation 3(3x+4)^2+18=27 are x=(sqrt(3)-4)/3 ,x=(-sqrt(3)-4)/3
Find the zeros of 3(3x+4)^2+18=27
The zeros of 3(3x+4)^2+18=27 are x=(sqrt(3)-4)/3 ,x=(-sqrt(3)-4)/3