What are the solutions to the equation 20=6+31t-16t^2 ?

The solutions to the equation 20=6+31t-16t^2 are t=(31-sqrt(65))/(32),t=(31+sqrt(65))/(32)

Find the zeros of 20=6+31t-16t^2

The zeros of 20=6+31t-16t^2 are t=(31-sqrt(65))/(32),t=(31+sqrt(65))/(32)

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20 = 6 + 31 t - 16 t^{2}

t = \frac{31 - 65}{32}, t = \frac{31 + 65}{32}

Decimal Notation

t = 0.71680 \dots, t = 1.22069 \dots

Solution steps

20 = 6 + 31 t - 16 t^{2}

Switch sides

6 + 31 t - 16 t^{2} = 20

Move

20

to the left side

- 16 t^{2} + 31 t - 14 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 31 \pm 3 1 ^{2} - 4 ( - 16 ) ( - 14 )}{2 ( - 16 )}

3 1^{2} - 4 (- 16) (- 14) = 65

t_{1, 2} = \frac{- 31 \pm 65}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 31 + 65}{2 ( - 16 )}, t_{2} = \frac{- 31 - 65}{2 ( - 16 )}

t = \frac{- 31 + 65}{2 ( - 16 )} : \frac{31 - 65}{32}

t = \frac{- 31 - 65}{2 ( - 16 )} : \frac{31 + 65}{32}

The solutions to the quadratic equation are:

t = \frac{31 - 65}{32}, t = \frac{31 + 65}{32}

b^{2} - 7 b = 0

8 x^{2} - 9 = - 3 x

(2 y + 4)^{2} = 16

3 (x + 3)^{2} = 75

4 - 4 y - y^{2} = 0

What are the solutions to the equation 20=6+31t-16t^2 ?
The solutions to the equation 20=6+31t-16t^2 are t=(31-sqrt(65))/(32),t=(31+sqrt(65))/(32)
Find the zeros of 20=6+31t-16t^2
The zeros of 20=6+31t-16t^2 are t=(31-sqrt(65))/(32),t=(31+sqrt(65))/(32)