What are the solutions to the equation 0=75+29t-5t^2 ?

The solutions to the equation 0=75+29t-5t^2 are t=-(-29+sqrt(2341))/(10),t=(29+sqrt(2341))/(10)

Find the zeros of 0=75+29t-5t^2

The zeros of 0=75+29t-5t^2 are t=-(-29+sqrt(2341))/(10),t=(29+sqrt(2341))/(10)

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0 = 75 + 29 t - 5 t^{2}

t = - \frac{- 29 + 2341}{10}, t = \frac{29 + 2341}{10}

Decimal Notation

t = - 1.93838 \dots, t = 7.73838 \dots

Solution steps

0 = 75 + 29 t - 5 t^{2}

Switch sides

75 + 29 t - 5 t^{2} = 0

Write in the standard form

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

- 5 t^{2} + 29 t + 75 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 29 \pm 2 9 ^{2} - 4 ( - 5 ) \cdot 75}{2 ( - 5 )}

2 9^{2} - 4 (- 5) \cdot 75 = 2341

t_{1, 2} = \frac{- 29 \pm 2341}{2 ( - 5 )}

Separate the solutions

t_{1} = \frac{- 29 + 2341}{2 ( - 5 )}, t_{2} = \frac{- 29 - 2341}{2 ( - 5 )}

t = \frac{- 29 + 2341}{2 ( - 5 )} : - \frac{- 29 + 2341}{10}

t = \frac{- 29 - 2341}{2 ( - 5 )} : \frac{29 + 2341}{10}

The solutions to the quadratic equation are:

t = - \frac{- 29 + 2341}{10}, t = \frac{29 + 2341}{10}

0 = 49 x^{2} - 16

7 a^{2} - 300 = 43

5 y^{2} + 26 y = - 5

x^{2} + 28 x + 196 = 0

(x + 4)^{2} - 5 = 31

What are the solutions to the equation 0=75+29t-5t^2 ?
The solutions to the equation 0=75+29t-5t^2 are t=-(-29+sqrt(2341))/(10),t=(29+sqrt(2341))/(10)
Find the zeros of 0=75+29t-5t^2
The zeros of 0=75+29t-5t^2 are t=-(-29+sqrt(2341))/(10),t=(29+sqrt(2341))/(10)