What are the solutions to the equation -16t^2+40t-24=0 ?

The solutions to the equation -16t^2+40t-24=0 are t=1,t= 3/2

Find the zeros of -16t^2+40t-24=0

The zeros of -16t^2+40t-24=0 are t=1,t= 3/2

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- 16 t^{2} + 40 t - 24 = 0

t = 1, t = \frac{3}{2}

Decimal Notation

t = 1, t = 1.5

Solution steps

- 16 t^{2} + 40 t - 24 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 40 \pm 4 0 ^{2} - 4 ( - 16 ) ( - 24 )}{2 ( - 16 )}

4 0^{2} - 4 (- 16) (- 24) = 8

t_{1, 2} = \frac{- 40 \pm 8}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 40 + 8}{2 ( - 16 )}, t_{2} = \frac{- 40 - 8}{2 ( - 16 )}

t = \frac{- 40 + 8}{2 ( - 16 )} : 1

t = \frac{- 40 - 8}{2 ( - 16 )} : \frac{3}{2}

The solutions to the quadratic equation are:

t = 1, t = \frac{3}{2}

28 = (x + 3) (x)

2 m^{2} = - 200

a x^{2} + b x - 1 + x = 0

2 w^{2} - 11 w - 40 = 0

12 x^{2} + 37 x + 28 = 0

What are the solutions to the equation -16t^2+40t-24=0 ?
The solutions to the equation -16t^2+40t-24=0 are t=1,t= 3/2
Find the zeros of -16t^2+40t-24=0
The zeros of -16t^2+40t-24=0 are t=1,t= 3/2