What are the solutions to the equation -16t^2+12t+10=0 ?

The solutions to the equation -16t^2+12t+10=0 are t=-1/2 ,t= 5/4

Find the zeros of -16t^2+12t+10=0

The zeros of -16t^2+12t+10=0 are t=-1/2 ,t= 5/4

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- 16 t^{2} + 12 t + 10 = 0

t = - \frac{1}{2}, t = \frac{5}{4}

Decimal Notation

t = - 0.5, t = 1.25

Solution steps

- 16 t^{2} + 12 t + 10 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 ( - 16 ) \cdot 10}{2 ( - 16 )}

1 2^{2} - 4 (- 16) \cdot 10 = 28

t_{1, 2} = \frac{- 12 \pm 28}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 12 + 28}{2 ( - 16 )}, t_{2} = \frac{- 12 - 28}{2 ( - 16 )}

t = \frac{- 12 + 28}{2 ( - 16 )} : - \frac{1}{2}

t = \frac{- 12 - 28}{2 ( - 16 )} : \frac{5}{4}

The solutions to the quadratic equation are:

t = - \frac{1}{2}, t = \frac{5}{4}

5 (m - 1) - 7 m (m - 1)

49 x^{4} - 14 x^{3} - 28 x

- k + 3 k

(t - 6) (t + 6) = (t - 2) (t - 4)

\frac{1}{16 a} + \frac{5}{12 b} - \frac{9}{10 b ^{3}}

What are the solutions to the equation -16t^2+12t+10=0 ?
The solutions to the equation -16t^2+12t+10=0 are t=-1/2 ,t= 5/4
Find the zeros of -16t^2+12t+10=0
The zeros of -16t^2+12t+10=0 are t=-1/2 ,t= 5/4