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

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

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

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

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

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

Solution steps

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}

x^{2} + 18 x = - 79

1 x^{2} + 4 x + 3 = 0

(x + 4) (x - 6) = 0

3 x^{2} = - 12 x

x, f = (x + y)^{2}

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