What are the solutions to the equation 24=6+35t-16t^2 ?

The solutions to the equation 24=6+35t-16t^2 are t=(35-sqrt(73))/(32),t=(35+sqrt(73))/(32)

Find the zeros of 24=6+35t-16t^2

The zeros of 24=6+35t-16t^2 are t=(35-sqrt(73))/(32),t=(35+sqrt(73))/(32)

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24 = 6 + 35 t - 16 t^{2}

t = \frac{35 - 73}{32}, t = \frac{35 + 73}{32}

Decimal Notation

t = 0.82674 \dots, t = 1.36075 \dots

Solution steps

24 = 6 + 35 t - 16 t^{2}

Switch sides

6 + 35 t - 16 t^{2} = 24

Move

24

to the left side

- 16 t^{2} + 35 t - 18 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 35 \pm 3 5 ^{2} - 4 ( - 16 ) ( - 18 )}{2 ( - 16 )}

3 5^{2} - 4 (- 16) (- 18) = 73

t_{1, 2} = \frac{- 35 \pm 73}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 35 + 73}{2 ( - 16 )}, t_{2} = \frac{- 35 - 73}{2 ( - 16 )}

t = \frac{- 35 + 73}{2 ( - 16 )} : \frac{35 - 73}{32}

t = \frac{- 35 - 73}{2 ( - 16 )} : \frac{35 + 73}{32}

The solutions to the quadratic equation are:

t = \frac{35 - 73}{32}, t = \frac{35 + 73}{32}

\frac{6 ( - 2 )}{3 ( - 2 ) ^{2} - ( 2 ) ^{3} - 3}

1^{\frac{1}{3}}

x \leq 1 and x \geq 3

2 w^{2} + 3 w - 35

12 t - 10 = 14 t + 2

What are the solutions to the equation 24=6+35t-16t^2 ?
The solutions to the equation 24=6+35t-16t^2 are t=(35-sqrt(73))/(32),t=(35+sqrt(73))/(32)
Find the zeros of 24=6+35t-16t^2
The zeros of 24=6+35t-16t^2 are t=(35-sqrt(73))/(32),t=(35+sqrt(73))/(32)