What are the solutions to the equation 12y^2-17y+6=0 ?

The solutions to the equation 12y^2-17y+6=0 are y= 3/4 ,y= 2/3

Find the zeros of 12y^2-17y+6=0

The zeros of 12y^2-17y+6=0 are y= 3/4 ,y= 2/3

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12 y^{2} - 17 y + 6 = 0

y = \frac{3}{4}, y = \frac{2}{3}

Decimal Notation

y = 0.75, y = 0.66666 \dots

Solution steps

12 y^{2} - 17 y + 6 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 17 ) \pm ( - 17 ) ^{2} - 4 \cdot 12 \cdot 6}{2 \cdot 12}

(- 17)^{2} - 4 \cdot 12 \cdot 6 = 1

y_{1, 2} = \frac{- ( - 17 ) \pm 1}{2 \cdot 12}

Separate the solutions

y_{1} = \frac{- ( - 17 ) + 1}{2 \cdot 12}, y_{2} = \frac{- ( - 17 ) - 1}{2 \cdot 12}

y = \frac{- ( - 17 ) + 1}{2 \cdot 12} : \frac{3}{4}

y = \frac{- ( - 17 ) - 1}{2 \cdot 12} : \frac{2}{3}

The solutions to the quadratic equation are:

y = \frac{3}{4}, y = \frac{2}{3}

x^{2} + 16 x + 9 = 20

x, x^{2} + y^{2} - x y = 1

(x + 1) (2 x - 3) = 5

x^{2} + \frac{1}{2} x - 3 = 0

x, f = x^{2} - 6 x + m - 3

What are the solutions to the equation 12y^2-17y+6=0 ?
The solutions to the equation 12y^2-17y+6=0 are y= 3/4 ,y= 2/3
Find the zeros of 12y^2-17y+6=0
The zeros of 12y^2-17y+6=0 are y= 3/4 ,y= 2/3