What are the solutions to the equation 0=-16t^2+6t+5 ?

The solutions to the equation 0=-16t^2+6t+5 are t=-(-3+sqrt(89))/(16),t=(3+sqrt(89))/(16)

Find the zeros of 0=-16t^2+6t+5

The zeros of 0=-16t^2+6t+5 are t=-(-3+sqrt(89))/(16),t=(3+sqrt(89))/(16)

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0 = - 16 t^{2} + 6 t + 5

t = - \frac{- 3 + 89}{16}, t = \frac{3 + 89}{16}

Decimal Notation

t = - 0.40212 \dots, t = 0.77712 \dots

Solution steps

0 = - 16 t^{2} + 6 t + 5

Switch sides

- 16 t^{2} + 6 t + 5 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 ( - 16 ) \cdot 5}{2 ( - 16 )}

6^{2} - 4 (- 16) \cdot 5 = 289

t_{1, 2} = \frac{- 6 \pm 2 89}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 6 + 2 89}{2 ( - 16 )}, t_{2} = \frac{- 6 - 2 89}{2 ( - 16 )}

t = \frac{- 6 + 2 89}{2 ( - 16 )} : - \frac{- 3 + 89}{16}

t = \frac{- 6 - 2 89}{2 ( - 16 )} : \frac{3 + 89}{16}

The solutions to the quadratic equation are:

t = - \frac{- 3 + 89}{16}, t = \frac{3 + 89}{16}

b^{2} + 8 b + 7

3 (x - 1) = 6 x

\frac{2 x + 2}{3} \leq 2

\frac{3 x - 2 y}{9 x ^{2} - 4 y ^{2}}

x^{2} < 49

What are the solutions to the equation 0=-16t^2+6t+5 ?
The solutions to the equation 0=-16t^2+6t+5 are t=-(-3+sqrt(89))/(16),t=(3+sqrt(89))/(16)
Find the zeros of 0=-16t^2+6t+5
The zeros of 0=-16t^2+6t+5 are t=-(-3+sqrt(89))/(16),t=(3+sqrt(89))/(16)