What are the solutions to the equation 15=40t-5t^2 ?

The solutions to the equation 15=40t-5t^2 are t=4-sqrt(13),t=4+sqrt(13)

Find the zeros of 15=40t-5t^2

The zeros of 15=40t-5t^2 are t=4-sqrt(13),t=4+sqrt(13)

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15 = 40 t - 5 t^{2}

t = 4 - 13, t = 4 + 13

Decimal Notation

t = 0.39444 \dots, t = 7.60555 \dots

Solution steps

15 = 40 t - 5 t^{2}

Switch sides

40 t - 5 t^{2} = 15

Move

15

to the left side

40 t - 5 t^{2} - 15 = 0

Write in the standard form

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

- 5 t^{2} + 40 t - 15 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 40 \pm 4 0 ^{2} - 4 ( - 5 ) ( - 15 )}{2 ( - 5 )}

4 0^{2} - 4 (- 5) (- 15) = 1013

t_{1, 2} = \frac{- 40 \pm 10 13}{2 ( - 5 )}

Separate the solutions

t_{1} = \frac{- 40 + 10 13}{2 ( - 5 )}, t_{2} = \frac{- 40 - 10 13}{2 ( - 5 )}

t = \frac{- 40 + 10 13}{2 ( - 5 )} : 4 - 13

t = \frac{- 40 - 10 13}{2 ( - 5 )} : 4 + 13

The solutions to the quadratic equation are:

t = 4 - 13, t = 4 + 13

1 0^{\circ} 2

2 (2 d + 3) < 34

\frac{4}{1 - i} + \frac{2 - i}{1 + i}

x^{3} - 6 x^{2} - 4 x + 24

418 - 152

What are the solutions to the equation 15=40t-5t^2 ?
The solutions to the equation 15=40t-5t^2 are t=4-sqrt(13),t=4+sqrt(13)
Find the zeros of 15=40t-5t^2
The zeros of 15=40t-5t^2 are t=4-sqrt(13),t=4+sqrt(13)