What are the solutions to the equation 8t^2+5t-5=0 ?

The solutions to the equation 8t^2+5t-5=0 are t=(-5+sqrt(185))/(16),t=(-5-sqrt(185))/(16)

Find the zeros of 8t^2+5t-5=0

The zeros of 8t^2+5t-5=0 are t=(-5+sqrt(185))/(16),t=(-5-sqrt(185))/(16)

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8 t^{2} + 5 t - 5 = 0

t = \frac{- 5 + 185}{16}, t = \frac{- 5 - 185}{16}

Decimal Notation

t = 0.53759 \dots, t = - 1.16259 \dots

Solution steps

8 t^{2} + 5 t - 5 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 8 ( - 5 )}{2 \cdot 8}

5^{2} - 4 \cdot 8 (- 5) = 185

t_{1, 2} = \frac{- 5 \pm 185}{2 \cdot 8}

Separate the solutions

t_{1} = \frac{- 5 + 185}{2 \cdot 8}, t_{2} = \frac{- 5 - 185}{2 \cdot 8}

t = \frac{- 5 + 185}{2 \cdot 8} : \frac{- 5 + 185}{16}

t = \frac{- 5 - 185}{2 \cdot 8} : \frac{- 5 - 185}{16}

The solutions to the quadratic equation are:

t = \frac{- 5 + 185}{16}, t = \frac{- 5 - 185}{16}

(- 32)^{\frac{6}{5}}

\frac{x ^{\frac{3}{4}}}{x ^{\frac{1}{2}}}

\frac{x + 2}{x + 7} < 2

x^{3} < 81

\frac{6}{8} = \frac{w}{12}

What are the solutions to the equation 8t^2+5t-5=0 ?
The solutions to the equation 8t^2+5t-5=0 are t=(-5+sqrt(185))/(16),t=(-5-sqrt(185))/(16)
Find the zeros of 8t^2+5t-5=0
The zeros of 8t^2+5t-5=0 are t=(-5+sqrt(185))/(16),t=(-5-sqrt(185))/(16)