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

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

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

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

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64 = 80 t - 16 t^{2}

t = 1, t = 4

Solution steps

64 = 80 t - 16 t^{2}

Switch sides

80 t - 16 t^{2} = 64

Move

64

to the left side

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

Write in the standard form

a x^{2} + b x + c = 0

- 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

2 x^{2} + x = - 9

x^{2} + 9 b^{2} = 0, b \neq = 0

X^{2} + 5 X + 2 = - 4

- 4 x^{2} - 24 x - 40

3 x^{2} - 18 x + 17

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