What are the solutions to the equation (3x-1)^2+12x-4=0 ?

The solutions to the equation (3x-1)^2+12x-4=0 are x= 1/3 ,x=-1

Find the zeros of (3x-1)^2+12x-4=0

The zeros of (3x-1)^2+12x-4=0 are x= 1/3 ,x=-1

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(3 x - 1)^{2} + 12 x - 4 = 0

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

Decimal Notation

x = 0.33333 \dots, x = - 1

Solution steps

(3 x - 1)^{2} + 12 x - 4 = 0

Expand

(3 x - 1)^{2} + 12 x - 4 : 9 x^{2} + 6 x - 3

9 x^{2} + 6 x - 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 9 ( - 3 )}{2 \cdot 9}

6^{2} - 4 \cdot 9 (- 3) = 12

x_{1, 2} = \frac{- 6 \pm 12}{2 \cdot 9}

Separate the solutions

x_{1} = \frac{- 6 + 12}{2 \cdot 9}, x_{2} = \frac{- 6 - 12}{2 \cdot 9}

x = \frac{- 6 + 12}{2 \cdot 9} : \frac{1}{3}

x = \frac{- 6 - 12}{2 \cdot 9} : - 1

The solutions to the quadratic equation are:

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

\frac{4 ^{9}}{4 ^{5}}

\frac{3}{4} x + 1 < 4

\frac{1}{2} (- 2)^{2} - 6 \times \frac{2}{3} + 6

25 x^{2} + 40 x y + 16 y^{2}

40 - 2 y = 30

What are the solutions to the equation (3x-1)^2+12x-4=0 ?
The solutions to the equation (3x-1)^2+12x-4=0 are x= 1/3 ,x=-1
Find the zeros of (3x-1)^2+12x-4=0
The zeros of (3x-1)^2+12x-4=0 are x= 1/3 ,x=-1