What are the solutions to the equation 136y^2+36y-681=0 ?

The solutions to the equation 136y^2+36y-681=0 are y=(-9+sqrt(23235))/(68),y=-(9+sqrt(23235))/(68)

Find the zeros of 136y^2+36y-681=0

The zeros of 136y^2+36y-681=0 are y=(-9+sqrt(23235))/(68),y=-(9+sqrt(23235))/(68)

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136 y^{2} + 36 y - 681 = 0

y = \frac{- 9 + 23235}{68}, y = - \frac{9 + 23235}{68}

Decimal Notation

y = 2.10926 \dots, y = - 2.37397 \dots

Solution steps

136 y^{2} + 36 y - 681 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 36 \pm 3 6 ^{2} - 4 \cdot 136 ( - 681 )}{2 \cdot 136}

3 6^{2} - 4 \cdot 136 (- 681) = 423235

y_{1, 2} = \frac{- 36 \pm 4 23235}{2 \cdot 136}

Separate the solutions

y_{1} = \frac{- 36 + 4 23235}{2 \cdot 136}, y_{2} = \frac{- 36 - 4 23235}{2 \cdot 136}

y = \frac{- 36 + 4 23235}{2 \cdot 136} : \frac{- 9 + 23235}{68}

y = \frac{- 36 - 4 23235}{2 \cdot 136} : - \frac{9 + 23235}{68}

The solutions to the quadratic equation are:

y = \frac{- 9 + 23235}{68}, y = - \frac{9 + 23235}{68}

5 t^{2} - 2 t - 1 = 0

\frac{9 x ^{2} + 3 x}{4 + 2} = 0

\frac{x ^{2}}{2 + 2} = 20

\frac{0.44 x ( x + x \cdot 5 - \frac{23}{55} ) + 3}{14 \cdot 2.24} = 1

48 x^{2} - 13 x - 1 = 0

What are the solutions to the equation 136y^2+36y-681=0 ?
The solutions to the equation 136y^2+36y-681=0 are y=(-9+sqrt(23235))/(68),y=-(9+sqrt(23235))/(68)
Find the zeros of 136y^2+36y-681=0
The zeros of 136y^2+36y-681=0 are y=(-9+sqrt(23235))/(68),y=-(9+sqrt(23235))/(68)