What are the solutions to the equation 15y^2-4y-4=0 ?

The solutions to the equation 15y^2-4y-4=0 are y= 2/3 ,y=-2/5

Find the zeros of 15y^2-4y-4=0

The zeros of 15y^2-4y-4=0 are y= 2/3 ,y=-2/5

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15 y^{2} - 4 y - 4 = 0

y = \frac{2}{3}, y = - \frac{2}{5}

Decimal Notation

y = 0.66666 \dots, y = - 0.4

Solution steps

15 y^{2} - 4 y - 4 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 15 ( - 4 )}{2 \cdot 15}

(- 4)^{2} - 4 \cdot 15 (- 4) = 16

y_{1, 2} = \frac{- ( - 4 ) \pm 16}{2 \cdot 15}

Separate the solutions

y_{1} = \frac{- ( - 4 ) + 16}{2 \cdot 15}, y_{2} = \frac{- ( - 4 ) - 16}{2 \cdot 15}

y = \frac{- ( - 4 ) + 16}{2 \cdot 15} : \frac{2}{3}

y = \frac{- ( - 4 ) - 16}{2 \cdot 15} : - \frac{2}{5}

The solutions to the quadratic equation are:

y = \frac{2}{3}, y = - \frac{2}{5}

2 n^{2} - 19 n + 17 = 3 n^{2}

x^{2} + (x - 5)^{2} = 73

x^{2} = - 6, x^{1} = 1

2 y^{2} + y - 11 = (y + 1)^{2}

8 x^{2} - 2 x = 0

What are the solutions to the equation 15y^2-4y-4=0 ?
The solutions to the equation 15y^2-4y-4=0 are y= 2/3 ,y=-2/5
Find the zeros of 15y^2-4y-4=0
The zeros of 15y^2-4y-4=0 are y= 2/3 ,y=-2/5