What are the solutions to the equation 520=-16t^2+108t+520 ?

The solutions to the equation 520=-16t^2+108t+520 are t=0,t= 27/4

Find the zeros of 520=-16t^2+108t+520

The zeros of 520=-16t^2+108t+520 are t=0,t= 27/4

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520 = - 16 t^{2} + 108 t + 520

t = 0, t = \frac{27}{4}

Decimal Notation

t = 0, t = 6.75

Solution steps

520 = - 16 t^{2} + 108 t + 520

Switch sides

- 16 t^{2} + 108 t + 520 = 520

Move

520

to the left side

- 16 t^{2} + 108 t = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 108 \pm 10 8 ^{2} - 4 ( - 16 ) \cdot 0}{2 ( - 16 )}

10 8^{2} - 4 (- 16) \cdot 0 = 108

t_{1, 2} = \frac{- 108 \pm 108}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 108 + 108}{2 ( - 16 )}, t_{2} = \frac{- 108 - 108}{2 ( - 16 )}

t = \frac{- 108 + 108}{2 ( - 16 )} : 0

t = \frac{- 108 - 108}{2 ( - 16 )} : \frac{27}{4}

The solutions to the quadratic equation are:

t = 0, t = \frac{27}{4}

5 X - 12 > 3 X - 4

28 - 32 x = 92

- 8 n \cdot 11 n^{9}

6 = \frac{x}{5}

3 y + 6 = 5 y - 18

What are the solutions to the equation 520=-16t^2+108t+520 ?
The solutions to the equation 520=-16t^2+108t+520 are t=0,t= 27/4
Find the zeros of 520=-16t^2+108t+520
The zeros of 520=-16t^2+108t+520 are t=0,t= 27/4