What are the solutions to the equation 3x^2+(4(3x+1))/3 =0 ?

The solutions to the equation 3x^2+(4(3x+1))/3 =0 are x=-2/3

Find the zeros of 3x^2+(4(3x+1))/3 =0

The zeros of 3x^2+(4(3x+1))/3 =0 are x=-2/3

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3 x^{2} + \frac{4 ( 3 x + 1 )}{3} = 0

x = - \frac{2}{3}

Decimal

x = - 0.66666 \dots

Solution steps

3 x^{2} + \frac{4 ( 3 x + 1 )}{3} = 0

Multiply both sides by

3

9 x^{2} + 4 (3 x + 1) = 0

Expand

9 x^{2} + 4 (3 x + 1) : 9 x^{2} + 12 x + 4

9 x^{2} + 12 x + 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 9 \cdot 4}{2 \cdot 9}

1 2^{2} - 4 \cdot 9 \cdot 4 = 0

x_{1, 2} = \frac{- 12 \pm 0}{2 \cdot 9}

x = \frac{- 12}{2 \cdot 9}

\frac{- 12}{2 \cdot 9} = - \frac{2}{3}

x = - \frac{2}{3}

The solution to the quadratic equation is:

x = - \frac{2}{3}

5 n^{2} - 6 n - 27 = 0

x^{2} - a^{2} - b x - ab = 0

(x + 4) (x - 2) = 8

(x - 1)^{2} + 15 = 14

- 2 x^{2} - 12 x - 16 = 0

What are the solutions to the equation 3x^2+(4(3x+1))/3 =0 ?
The solutions to the equation 3x^2+(4(3x+1))/3 =0 are x=-2/3
Find the zeros of 3x^2+(4(3x+1))/3 =0
The zeros of 3x^2+(4(3x+1))/3 =0 are x=-2/3