What are the solutions to the equation 35=70t-5t^2 ?

The solutions to the equation 35=70t-5t^2 are t=7-sqrt(42),t=7+sqrt(42)

Find the zeros of 35=70t-5t^2

The zeros of 35=70t-5t^2 are t=7-sqrt(42),t=7+sqrt(42)

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35 = 70 t - 5 t^{2}

t = 7 - 42, t = 7 + 42

Decimal Notation

t = 0.51925 \dots, t = 13.48074 \dots

Solution steps

35 = 70 t - 5 t^{2}

Switch sides

70 t - 5 t^{2} = 35

Move

35

to the left side

70 t - 5 t^{2} - 35 = 0

Write in the standard form

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

- 5 t^{2} + 70 t - 35 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 70 \pm 7 0 ^{2} - 4 ( - 5 ) ( - 35 )}{2 ( - 5 )}

7 0^{2} - 4 (- 5) (- 35) = 1042

t_{1, 2} = \frac{- 70 \pm 10 42}{2 ( - 5 )}

Separate the solutions

t_{1} = \frac{- 70 + 10 42}{2 ( - 5 )}, t_{2} = \frac{- 70 - 10 42}{2 ( - 5 )}

t = \frac{- 70 + 10 42}{2 ( - 5 )} : 7 - 42

t = \frac{- 70 - 10 42}{2 ( - 5 )} : 7 + 42

The solutions to the quadratic equation are:

t = 7 - 42, t = 7 + 42

5 x = 7 (x + 8)

4 X^{2} - 4 X - 8 = - 9

(3 x + 2)^{2} = (x + 2) (x - 3)

- 2 x + 12 < 48

x - 17 = 23

What are the solutions to the equation 35=70t-5t^2 ?
The solutions to the equation 35=70t-5t^2 are t=7-sqrt(42),t=7+sqrt(42)
Find the zeros of 35=70t-5t^2
The zeros of 35=70t-5t^2 are t=7-sqrt(42),t=7+sqrt(42)