What are the solutions to the equation -16t^2+10t+6=0 ?

The solutions to the equation -16t^2+10t+6=0 are t=-3/8 ,t=1

Find the zeros of -16t^2+10t+6=0

The zeros of -16t^2+10t+6=0 are t=-3/8 ,t=1

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- 16 t^{2} + 10 t + 6 = 0

t = - \frac{3}{8}, t = 1

Decimal Notation

t = - 0.375, t = 1

Solution steps

- 16 t^{2} + 10 t + 6 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 ( - 16 ) \cdot 6}{2 ( - 16 )}

1 0^{2} - 4 (- 16) \cdot 6 = 22

t_{1, 2} = \frac{- 10 \pm 22}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 10 + 22}{2 ( - 16 )}, t_{2} = \frac{- 10 - 22}{2 ( - 16 )}

t = \frac{- 10 + 22}{2 ( - 16 )} : - \frac{3}{8}

t = \frac{- 10 - 22}{2 ( - 16 )} : 1

The solutions to the quadratic equation are:

t = - \frac{3}{8}, t = 1

x, 9 x^{2} - 4 y^{2} + 36 x + 24 y = 0

2 x^{2} + 8 x + 4

x, x^{2} - y^{2} = a

3 x^{2} + 12 x + 25 = 0

w^{2} + 10 w - 3 = 0

What are the solutions to the equation -16t^2+10t+6=0 ?
The solutions to the equation -16t^2+10t+6=0 are t=-3/8 ,t=1
Find the zeros of -16t^2+10t+6=0
The zeros of -16t^2+10t+6=0 are t=-3/8 ,t=1