What are the solutions to the equation -12t^2+122t-109=0 ?

The solutions to the equation -12t^2+122t-109=0 are t=(61-sqrt(2413))/(12),t=(61+sqrt(2413))/(12)

Find the zeros of -12t^2+122t-109=0

The zeros of -12t^2+122t-109=0 are t=(61-sqrt(2413))/(12),t=(61+sqrt(2413))/(12)

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- 12 t^{2} + 122 t - 109 = 0

t = \frac{61 - 2413}{12}, t = \frac{61 + 2413}{12}

Decimal

t = 0.98980 \dots, t = 9.17685 \dots

Solution steps

- 12 t^{2} + 122 t - 109 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 122 \pm 12 2 ^{2} - 4 ( - 12 ) ( - 109 )}{2 ( - 12 )}

12 2^{2} - 4 (- 12) (- 109) = 22413

t_{1, 2} = \frac{- 122 \pm 2 2413}{2 ( - 12 )}

Separate the solutions

t_{1} = \frac{- 122 + 2 2413}{2 ( - 12 )}, t_{2} = \frac{- 122 - 2 2413}{2 ( - 12 )}

t = \frac{- 122 + 2 2413}{2 ( - 12 )} : \frac{61 - 2413}{12}

t = \frac{- 122 - 2 2413}{2 ( - 12 )} : \frac{61 + 2413}{12}

The solutions to the quadratic equation are:

t = \frac{61 - 2413}{12}, t = \frac{61 + 2413}{12}

5 c^{2} + 12 c + 4 = 0

2 (x - 5)^{2} + 2 = 20

8 x^{2} = 72 x

(3 x - 2)^{2} = (x + 1)^{2}

y, x^{2} - 14 x y + y^{2} = 14

What are the solutions to the equation -12t^2+122t-109=0 ?
The solutions to the equation -12t^2+122t-109=0 are t=(61-sqrt(2413))/(12),t=(61+sqrt(2413))/(12)
Find the zeros of -12t^2+122t-109=0
The zeros of -12t^2+122t-109=0 are t=(61-sqrt(2413))/(12),t=(61+sqrt(2413))/(12)