What are the solutions to the equation 12y^2-17y-5=0 ?

The solutions to the equation 12y^2-17y-5=0 are y= 5/3 ,y=-1/4

Find the zeros of 12y^2-17y-5=0

The zeros of 12y^2-17y-5=0 are y= 5/3 ,y=-1/4

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12 y^{2} - 17 y - 5 = 0

y = \frac{5}{3}, y = - \frac{1}{4}

Decimal Notation

y = 1.66666 \dots, y = - 0.25

Solution steps

12 y^{2} - 17 y - 5 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 17 ) \pm ( - 17 ) ^{2} - 4 \cdot 12 ( - 5 )}{2 \cdot 12}

(- 17)^{2} - 4 \cdot 12 (- 5) = 23

y_{1, 2} = \frac{- ( - 17 ) \pm 23}{2 \cdot 12}

Separate the solutions

y_{1} = \frac{- ( - 17 ) + 23}{2 \cdot 12}, y_{2} = \frac{- ( - 17 ) - 23}{2 \cdot 12}

y = \frac{- ( - 17 ) + 23}{2 \cdot 12} : \frac{5}{3}

y = \frac{- ( - 17 ) - 23}{2 \cdot 12} : - \frac{1}{4}

The solutions to the quadratic equation are:

y = \frac{5}{3}, y = - \frac{1}{4}

x^{2} + 8 x + 15 = - 4

12 x^{2} - 108 x + 216 = 0

3 x^{2} - x = 1

2 r^{2} + 9 r + 10 = 0

A, (A + B) (A + C) = A + BC

What are the solutions to the equation 12y^2-17y-5=0 ?
The solutions to the equation 12y^2-17y-5=0 are y= 5/3 ,y=-1/4
Find the zeros of 12y^2-17y-5=0
The zeros of 12y^2-17y-5=0 are y= 5/3 ,y=-1/4