What are the solutions to the equation 0=-5t^2+25t ?

The solutions to the equation 0=-5t^2+25t are t=0,t=5

Find the zeros of 0=-5t^2+25t

The zeros of 0=-5t^2+25t are t=0,t=5

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0 = - 5 t^{2} + 25 t

t = 0, t = 5

Solution steps

0 = - 5 t^{2} + 25 t

Switch sides

- 5 t^{2} + 25 t = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 25 \pm 2 5 ^{2} - 4 ( - 5 ) \cdot 0}{2 ( - 5 )}

2 5^{2} - 4 (- 5) \cdot 0 = 25

t_{1, 2} = \frac{- 25 \pm 25}{2 ( - 5 )}

Separate the solutions

t_{1} = \frac{- 25 + 25}{2 ( - 5 )}, t_{2} = \frac{- 25 - 25}{2 ( - 5 )}

t = \frac{- 25 + 25}{2 ( - 5 )} : 0

t = \frac{- 25 - 25}{2 ( - 5 )} : 5

The solutions to the quadratic equation are:

t = 0, t = 5

2 x \cdot x^{2}

14000 = 0.125 x^{2} + 20 x + 5000

a - 5 = - 12

(x - 7)^{2} - 9 = 0

(185) (130 - x) = (266.67) (45 - x)^{2}

What are the solutions to the equation 0=-5t^2+25t ?
The solutions to the equation 0=-5t^2+25t are t=0,t=5
Find the zeros of 0=-5t^2+25t
The zeros of 0=-5t^2+25t are t=0,t=5