What are the solutions to the equation 16=3+40t-16t^2 ?

The solutions to the equation 16=3+40t-16t^2 are t=(5-2sqrt(3))/4 ,t=(5+2sqrt(3))/4

Find the zeros of 16=3+40t-16t^2

The zeros of 16=3+40t-16t^2 are t=(5-2sqrt(3))/4 ,t=(5+2sqrt(3))/4

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16 = 3 + 40 t - 16 t^{2}

t = \frac{5 - 2 3}{4}, t = \frac{5 + 2 3}{4}

Decimal Notation

t = 0.38397 \dots, t = 2.11602 \dots

Solution steps

16 = 3 + 40 t - 16 t^{2}

Switch sides

3 + 40 t - 16 t^{2} = 16

Move

16

to the left side

- 16 t^{2} + 40 t - 13 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 40 \pm 4 0 ^{2} - 4 ( - 16 ) ( - 13 )}{2 ( - 16 )}

4 0^{2} - 4 (- 16) (- 13) = 163

t_{1, 2} = \frac{- 40 \pm 16 3}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 40 + 16 3}{2 ( - 16 )}, t_{2} = \frac{- 40 - 16 3}{2 ( - 16 )}

t = \frac{- 40 + 16 3}{2 ( - 16 )} : \frac{5 - 2 3}{4}

t = \frac{- 40 - 16 3}{2 ( - 16 )} : \frac{5 + 2 3}{4}

The solutions to the quadratic equation are:

t = \frac{5 - 2 3}{4}, t = \frac{5 + 2 3}{4}

- 4 x + 7 \leq 15

3 x^{2} = 9 x + 12

13 = ∣ 2 - 3 x ∣

2 (5 w + 9) \leq 7 w + 3

∣ x + 6 ∣ > 10

What are the solutions to the equation 16=3+40t-16t^2 ?
The solutions to the equation 16=3+40t-16t^2 are t=(5-2sqrt(3))/4 ,t=(5+2sqrt(3))/4
Find the zeros of 16=3+40t-16t^2
The zeros of 16=3+40t-16t^2 are t=(5-2sqrt(3))/4 ,t=(5+2sqrt(3))/4