What are the solutions to the equation 12x^2-42x+28=0 ?

The solutions to the equation 12x^2-42x+28=0 are x=(21+sqrt(105))/(12),x=(21-sqrt(105))/(12)

Find the zeros of 12x^2-42x+28=0

The zeros of 12x^2-42x+28=0 are x=(21+sqrt(105))/(12),x=(21-sqrt(105))/(12)

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12 x^{2} - 42 x + 28 = 0

x = \frac{21 + 105}{12}, x = \frac{21 - 105}{12}

Decimal Notation

x = 2.60391 \dots, x = 0.89608 \dots

Solution steps

12 x^{2} - 42 x + 28 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 42 ) \pm ( - 42 ) ^{2} - 4 \cdot 12 \cdot 28}{2 \cdot 12}

(- 42)^{2} - 4 \cdot 12 \cdot 28 = 2105

x_{1, 2} = \frac{- ( - 42 ) \pm 2 105}{2 \cdot 12}

Separate the solutions

x_{1} = \frac{- ( - 42 ) + 2 105}{2 \cdot 12}, x_{2} = \frac{- ( - 42 ) - 2 105}{2 \cdot 12}

x = \frac{- ( - 42 ) + 2 105}{2 \cdot 12} : \frac{21 + 105}{12}

x = \frac{- ( - 42 ) - 2 105}{2 \cdot 12} : \frac{21 - 105}{12}

The solutions to the quadratic equation are:

x = \frac{21 + 105}{12}, x = \frac{21 - 105}{12}

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

z^{2} + 1 = 2

- k^{2} - k + 2 = 0

9 r^{2} + 21 r + 9 = r^{2}

(2 x^{2} + 3) = 8 x + 3

What are the solutions to the equation 12x^2-42x+28=0 ?
The solutions to the equation 12x^2-42x+28=0 are x=(21+sqrt(105))/(12),x=(21-sqrt(105))/(12)
Find the zeros of 12x^2-42x+28=0
The zeros of 12x^2-42x+28=0 are x=(21+sqrt(105))/(12),x=(21-sqrt(105))/(12)