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

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

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

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

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

t = 2, t = 5

Solution steps

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

Switch sides

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

Move

160

to the left side

- 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 - y)^{2} - a \cdot (x - y)

y, 7 x^{3} - 3 x^{2} y + 2 y^{2} = 5 y + 2 x

2 x = 38

A^{2} = 4322.15

2 (p + 1) + 8 (p - 1) = 5 p

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