What are the solutions to the equation 60=-0.3t^2+9.7t-2 ?

The solutions to the equation 60=-0.3t^2+9.7t-2 are t=(97-sqrt(1969))/6 ,t=(97+sqrt(1969))/6

Find the zeros of 60=-0.3t^2+9.7t-2

The zeros of 60=-0.3t^2+9.7t-2 are t=(97-sqrt(1969))/6 ,t=(97+sqrt(1969))/6

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60 = - 0.3 t^{2} + 9.7 t - 2

t = \frac{97 - 1969}{6}, t = \frac{97 + 1969}{6}

Decimal Notation

t = 8.77109 \dots, t = 23.56223 \dots

Solution steps

60 = - 0.3 t^{2} + 9.7 t - 2

Multiply both sides by

10

600 = - 3 t^{2} + 97 t - 20

Switch sides

- 3 t^{2} + 97 t - 20 = 600

Move

600

to the left side

- 3 t^{2} + 97 t - 620 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 97 \pm 9 7 ^{2} - 4 ( - 3 ) ( - 620 )}{2 ( - 3 )}

9 7^{2} - 4 (- 3) (- 620) = 1969

t_{1, 2} = \frac{- 97 \pm 1969}{2 ( - 3 )}

Separate the solutions

t_{1} = \frac{- 97 + 1969}{2 ( - 3 )}, t_{2} = \frac{- 97 - 1969}{2 ( - 3 )}

t = \frac{- 97 + 1969}{2 ( - 3 )} : \frac{97 - 1969}{6}

t = \frac{- 97 - 1969}{2 ( - 3 )} : \frac{97 + 1969}{6}

The solutions to the quadratic equation are:

t = \frac{97 - 1969}{6}, t = \frac{97 + 1969}{6}

z^{2} (2 x - y) + 16 (y - 2 x)

ab c + ab c^{2}

k^{3} + 27

- \frac{25}{3} + 3

16 - 4 x^{2}

What are the solutions to the equation 60=-0.3t^2+9.7t-2 ?
The solutions to the equation 60=-0.3t^2+9.7t-2 are t=(97-sqrt(1969))/6 ,t=(97+sqrt(1969))/6
Find the zeros of 60=-0.3t^2+9.7t-2
The zeros of 60=-0.3t^2+9.7t-2 are t=(97-sqrt(1969))/6 ,t=(97+sqrt(1969))/6