What are the solutions to the equation -2t^2+3t+8=0 ?

The solutions to the equation -2t^2+3t+8=0 are t=-(-3+sqrt(73))/4 ,t=(3+sqrt(73))/4

Find the zeros of -2t^2+3t+8=0

The zeros of -2t^2+3t+8=0 are t=-(-3+sqrt(73))/4 ,t=(3+sqrt(73))/4

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- 2 t^{2} + 3 t + 8 = 0

t = - \frac{- 3 + 73}{4}, t = \frac{3 + 73}{4}

Decimal Notation

t = - 1.38600 \dots, t = 2.88600 \dots

Solution steps

- 2 t^{2} + 3 t + 8 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 ( - 2 ) \cdot 8}{2 ( - 2 )}

3^{2} - 4 (- 2) \cdot 8 = 73

t_{1, 2} = \frac{- 3 \pm 73}{2 ( - 2 )}

Separate the solutions

t_{1} = \frac{- 3 + 73}{2 ( - 2 )}, t_{2} = \frac{- 3 - 73}{2 ( - 2 )}

t = \frac{- 3 + 73}{2 ( - 2 )} : - \frac{- 3 + 73}{4}

t = \frac{- 3 - 73}{2 ( - 2 )} : \frac{3 + 73}{4}

The solutions to the quadratic equation are:

t = - \frac{- 3 + 73}{4}, t = \frac{3 + 73}{4}

3.2 = \frac{t}{8}

x^{2} - x + 0.5 = 0

\frac{x - 3}{x - 5} > 0

9 x^{2} + 6 x + 5 = 0

y^{- \frac{3}{2}}

What are the solutions to the equation -2t^2+3t+8=0 ?
The solutions to the equation -2t^2+3t+8=0 are t=-(-3+sqrt(73))/4 ,t=(3+sqrt(73))/4
Find the zeros of -2t^2+3t+8=0
The zeros of -2t^2+3t+8=0 are t=-(-3+sqrt(73))/4 ,t=(3+sqrt(73))/4