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

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

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

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

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

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

Decimal Notation

x = 1.76759 \dots, x = 0.56574 \dots

Solution steps

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

Write in the standard form

a x^{2} + b x + c = 0

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} - 37 x = 0

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

4 x^{2} + 4 x + 5

x^{2} - 25 x + 100 = 0

x^{2} - 0.4 x - 1.2 = 0

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