What are the solutions to the equation 20=6+38t-16t^2 ?

The solutions to the equation 20=6+38t-16t^2 are t=(19-sqrt(137))/(16),t=(19+sqrt(137))/(16)

Find the zeros of 20=6+38t-16t^2

The zeros of 20=6+38t-16t^2 are t=(19-sqrt(137))/(16),t=(19+sqrt(137))/(16)

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20 = 6 + 38 t - 16 t^{2}

t = \frac{19 - 137}{16}, t = \frac{19 + 137}{16}

Decimal Notation

t = 0.45595 \dots, t = 1.91904 \dots

Solution steps

20 = 6 + 38 t - 16 t^{2}

Switch sides

6 + 38 t - 16 t^{2} = 20

Move

20

to the left side

- 16 t^{2} + 38 t - 14 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 38 \pm 3 8 ^{2} - 4 ( - 16 ) ( - 14 )}{2 ( - 16 )}

3 8^{2} - 4 (- 16) (- 14) = 2137

t_{1, 2} = \frac{- 38 \pm 2 137}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 38 + 2 137}{2 ( - 16 )}, t_{2} = \frac{- 38 - 2 137}{2 ( - 16 )}

t = \frac{- 38 + 2 137}{2 ( - 16 )} : \frac{19 - 137}{16}

t = \frac{- 38 - 2 137}{2 ( - 16 )} : \frac{19 + 137}{16}

The solutions to the quadratic equation are:

t = \frac{19 - 137}{16}, t = \frac{19 + 137}{16}

lo g_{2} (\frac{t - 8}{2 t}) < lo g_{2} (3) - 7

x^{3} - 17 x - 4

x^{2} = 6 x - 15

- 4 x - 7 - 3 x + 4 = 25

0.36

What are the solutions to the equation 20=6+38t-16t^2 ?
The solutions to the equation 20=6+38t-16t^2 are t=(19-sqrt(137))/(16),t=(19+sqrt(137))/(16)
Find the zeros of 20=6+38t-16t^2
The zeros of 20=6+38t-16t^2 are t=(19-sqrt(137))/(16),t=(19+sqrt(137))/(16)