What are the solutions to the equation 0=35+30t-5t^2 ?

The solutions to the equation 0=35+30t-5t^2 are t=-1,t=7

Find the zeros of 0=35+30t-5t^2

The zeros of 0=35+30t-5t^2 are t=-1,t=7

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0 = 35 + 30 t - 5 t^{2}

t = - 1, t = 7

Solution steps

0 = 35 + 30 t - 5 t^{2}

Switch sides

35 + 30 t - 5 t^{2} = 0

Write in the standard form

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

- 5 t^{2} + 30 t + 35 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 30 \pm 3 0 ^{2} - 4 ( - 5 ) \cdot 35}{2 ( - 5 )}

3 0^{2} - 4 (- 5) \cdot 35 = 40

t_{1, 2} = \frac{- 30 \pm 40}{2 ( - 5 )}

Separate the solutions

t_{1} = \frac{- 30 + 40}{2 ( - 5 )}, t_{2} = \frac{- 30 - 40}{2 ( - 5 )}

t = \frac{- 30 + 40}{2 ( - 5 )} : - 1

t = \frac{- 30 - 40}{2 ( - 5 )} : 7

The solutions to the quadratic equation are:

t = - 1, t = 7

16 z^{2} + 40 z + 25 = 0

x^{2} - 8 x + 8 = - 4

x^{2} = 1 - (- \frac{5}{6})^{2}

x^{2} - 2 x - 5 = 30

b^{2} - 9 b + 20 = 0

What are the solutions to the equation 0=35+30t-5t^2 ?
The solutions to the equation 0=35+30t-5t^2 are t=-1,t=7
Find the zeros of 0=35+30t-5t^2
The zeros of 0=35+30t-5t^2 are t=-1,t=7