What are the solutions to the equation 0=12x^2-28x+12 ?

The solutions to the equation 0=12x^2-28x+12 are x=(7+sqrt(13))/6 ,x=(7-sqrt(13))/6

Find the zeros of 0=12x^2-28x+12

The zeros of 0=12x^2-28x+12 are x=(7+sqrt(13))/6 ,x=(7-sqrt(13))/6

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0 = 12 x^{2} - 28 x + 12

x = \frac{7 + 13}{6}, x = \frac{7 - 13}{6}

Decimal Notation

x = 1.76759 \dots, x = 0.56574 \dots

Solution steps

0 = 12 x^{2} - 28 x + 12

Switch sides

12 x^{2} - 28 x + 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 28 ) \pm ( - 28 ) ^{2} - 4 \cdot 12 \cdot 12}{2 \cdot 12}

(- 28)^{2} - 4 \cdot 12 \cdot 12 = 413

x_{1, 2} = \frac{- ( - 28 ) \pm 4 13}{2 \cdot 12}

Separate the solutions

x_{1} = \frac{- ( - 28 ) + 4 13}{2 \cdot 12}, x_{2} = \frac{- ( - 28 ) - 4 13}{2 \cdot 12}

x = \frac{- ( - 28 ) + 4 13}{2 \cdot 12} : \frac{7 + 13}{6}

x = \frac{- ( - 28 ) - 4 13}{2 \cdot 12} : \frac{7 - 13}{6}

The solutions to the quadratic equation are:

x = \frac{7 + 13}{6}, x = \frac{7 - 13}{6}

(4 - x) (1 - x) + 2 = 0

2 z^{2} + 2.5 z + 3.75 = 0

6 x^{2} = 29 x + 5

2 x (x - 3) = x^{2} - 9

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

What are the solutions to the equation 0=12x^2-28x+12 ?
The solutions to the equation 0=12x^2-28x+12 are x=(7+sqrt(13))/6 ,x=(7-sqrt(13))/6
Find the zeros of 0=12x^2-28x+12
The zeros of 0=12x^2-28x+12 are x=(7+sqrt(13))/6 ,x=(7-sqrt(13))/6