What are the solutions to the equation 3(3x+1)^2-43=-13 ?

The solutions to the equation 3(3x+1)^2-43=-13 are x=(sqrt(10)-1)/3 ,x=(-sqrt(10)-1)/3

Find the zeros of 3(3x+1)^2-43=-13

The zeros of 3(3x+1)^2-43=-13 are x=(sqrt(10)-1)/3 ,x=(-sqrt(10)-1)/3

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3 (3 x + 1)^{2} - 43 = - 13

x = \frac{10 - 1}{3}, x = \frac{- 10 - 1}{3}

Decimal Notation

x = 0.72075 \dots, x = - 1.38742 \dots

Solution steps

3 (3 x + 1)^{2} - 43 = - 13

Move

43

to the right side

3 (3 x + 1)^{2} = 30

Divide both sides by

3

(3 x + 1)^{2} = 10

For

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

the solutions are

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

Solve

3 x + 1 = 10 : x = \frac{10 - 1}{3}

Solve

3 x + 1 = - 10 : x = \frac{- 10 - 1}{3}

The solutions to the quadratic equation are:

x = \frac{10 - 1}{3}, x = \frac{- 10 - 1}{3}

4 x^{2} - 20 x = - 5

x, s in x + x cos x = 0

x^{2} - 6 x + 8

7 x^{2} + 6 = 48

15 b^{2} - 8 b - 3 = - 4

What are the solutions to the equation 3(3x+1)^2-43=-13 ?
The solutions to the equation 3(3x+1)^2-43=-13 are x=(sqrt(10)-1)/3 ,x=(-sqrt(10)-1)/3
Find the zeros of 3(3x+1)^2-43=-13
The zeros of 3(3x+1)^2-43=-13 are x=(sqrt(10)-1)/3 ,x=(-sqrt(10)-1)/3