What are the solutions to the equation 20=45t-5t^2 ?

The solutions to the equation 20=45t-5t^2 are t=(9-sqrt(65))/2 ,t=(9+sqrt(65))/2

Find the zeros of 20=45t-5t^2

The zeros of 20=45t-5t^2 are t=(9-sqrt(65))/2 ,t=(9+sqrt(65))/2

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20 = 45 t - 5 t^{2}

t = \frac{9 - 65}{2}, t = \frac{9 + 65}{2}

Decimal Notation

t = 0.46887 \dots, t = 8.53112 \dots

Solution steps

20 = 45 t - 5 t^{2}

Switch sides

45 t - 5 t^{2} = 20

Move

20

to the left side

45 t - 5 t^{2} - 20 = 0

Write in the standard form

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

- 5 t^{2} + 45 t - 20 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 45 \pm 4 5 ^{2} - 4 ( - 5 ) ( - 20 )}{2 ( - 5 )}

4 5^{2} - 4 (- 5) (- 20) = 565

t_{1, 2} = \frac{- 45 \pm 5 65}{2 ( - 5 )}

Separate the solutions

t_{1} = \frac{- 45 + 5 65}{2 ( - 5 )}, t_{2} = \frac{- 45 - 5 65}{2 ( - 5 )}

t = \frac{- 45 + 5 65}{2 ( - 5 )} : \frac{9 - 65}{2}

t = \frac{- 45 - 5 65}{2 ( - 5 )} : \frac{9 + 65}{2}

The solutions to the quadratic equation are:

t = \frac{9 - 65}{2}, t = \frac{9 + 65}{2}

6 x^{2} + 23 x + 7 = 0

2 x^{2} = 3 x - 5

y, x^{3} - 4 x y - y^{2} = 3

13 x^{2} - 5 x = 0

y, x = y^{2} - 3 y + 2

What are the solutions to the equation 20=45t-5t^2 ?
The solutions to the equation 20=45t-5t^2 are t=(9-sqrt(65))/2 ,t=(9+sqrt(65))/2
Find the zeros of 20=45t-5t^2
The zeros of 20=45t-5t^2 are t=(9-sqrt(65))/2 ,t=(9+sqrt(65))/2