What are the solutions to the equation -36y^2+36y+1=0 ?

The solutions to the equation -36y^2+36y+1=0 are y=-(-3+sqrt(10))/6 ,y=(3+sqrt(10))/6

Find the zeros of -36y^2+36y+1=0

The zeros of -36y^2+36y+1=0 are y=-(-3+sqrt(10))/6 ,y=(3+sqrt(10))/6

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- 36 y^{2} + 36 y + 1 = 0

y = - \frac{- 3 + 10}{6}, y = \frac{3 + 10}{6}

Decimal Notation

y = - 0.02704 \dots, y = 1.02704 \dots

Solution steps

- 36 y^{2} + 36 y + 1 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 36 \pm 3 6 ^{2} - 4 ( - 36 ) \cdot 1}{2 ( - 36 )}

3 6^{2} - 4 (- 36) \cdot 1 = 1210

y_{1, 2} = \frac{- 36 \pm 12 10}{2 ( - 36 )}

Separate the solutions

y_{1} = \frac{- 36 + 12 10}{2 ( - 36 )}, y_{2} = \frac{- 36 - 12 10}{2 ( - 36 )}

y = \frac{- 36 + 12 10}{2 ( - 36 )} : - \frac{- 3 + 10}{6}

y = \frac{- 36 - 12 10}{2 ( - 36 )} : \frac{3 + 10}{6}

The solutions to the quadratic equation are:

y = - \frac{- 3 + 10}{6}, y = \frac{3 + 10}{6}

y, x^{3} y^{2} - 5 x^{2} y + x = 1

6 x^{2} - x + 1 = x^{2} + x + 4

y, x^{2} - 6 x y + y^{2} = 6

9 x^{2} + 6 x + 4

x, z = 64 - x^{2} - y^{2}

What are the solutions to the equation -36y^2+36y+1=0 ?
The solutions to the equation -36y^2+36y+1=0 are y=-(-3+sqrt(10))/6 ,y=(3+sqrt(10))/6
Find the zeros of -36y^2+36y+1=0
The zeros of -36y^2+36y+1=0 are y=-(-3+sqrt(10))/6 ,y=(3+sqrt(10))/6