What are the solutions to the equation 64=-16t^2+80t ?

The solutions to the equation 64=-16t^2+80t are t=1,t=4

Find the zeros of 64=-16t^2+80t

The zeros of 64=-16t^2+80t are t=1,t=4

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64 = - 16 t^{2} + 80 t

t = 1, t = 4

Solution steps

64 = - 16 t^{2} + 80 t

Switch sides

- 16 t^{2} + 80 t = 64

Move

64

to the left side

- 16 t^{2} + 80 t - 64 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 80 \pm 8 0 ^{2} - 4 ( - 16 ) ( - 64 )}{2 ( - 16 )}

8 0^{2} - 4 (- 16) (- 64) = 48

t_{1, 2} = \frac{- 80 \pm 48}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 80 + 48}{2 ( - 16 )}, t_{2} = \frac{- 80 - 48}{2 ( - 16 )}

t = \frac{- 80 + 48}{2 ( - 16 )} : 1

t = \frac{- 80 - 48}{2 ( - 16 )} : 4

The solutions to the quadratic equation are:

t = 1, t = 4

x, 2 x^{2} - x y + y = 1

5 x^{2} - 50 x = - 60

4 = 3 y^{2} + 11 y

6 c^{2} + 9 = 5 c

8 x^{2} + 6 x = 8

What are the solutions to the equation 64=-16t^2+80t ?
The solutions to the equation 64=-16t^2+80t are t=1,t=4
Find the zeros of 64=-16t^2+80t
The zeros of 64=-16t^2+80t are t=1,t=4