What are the solutions to the equation -16t^2+112t-160=0 ?

The solutions to the equation -16t^2+112t-160=0 are t=2,t=5

Find the zeros of -16t^2+112t-160=0

The zeros of -16t^2+112t-160=0 are t=2,t=5

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- 16 t^{2} + 112 t - 160 = 0

t = 2, t = 5

Solution steps

- 16 t^{2} + 112 t - 160 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 112 \pm 11 2 ^{2} - 4 ( - 16 ) ( - 160 )}{2 ( - 16 )}

11 2^{2} - 4 (- 16) (- 160) = 48

t_{1, 2} = \frac{- 112 \pm 48}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 112 + 48}{2 ( - 16 )}, t_{2} = \frac{- 112 - 48}{2 ( - 16 )}

t = \frac{- 112 + 48}{2 ( - 16 )} : 2

t = \frac{- 112 - 48}{2 ( - 16 )} : 5

The solutions to the quadratic equation are:

t = 2, t = 5

x^{2} + 5 x - 19 = 0

4 x^{2} - 7 x + 9 = 0

- 2 (x - 4)^{2} + 50 = 0

x, y^{2} = x^{2} + z^{2}

2 x - x^{2} + 3 = 0

What are the solutions to the equation -16t^2+112t-160=0 ?
The solutions to the equation -16t^2+112t-160=0 are t=2,t=5
Find the zeros of -16t^2+112t-160=0
The zeros of -16t^2+112t-160=0 are t=2,t=5