What are the solutions to the equation 15=3+36t-16t^2 ?

The solutions to the equation 15=3+36t-16t^2 are t=(9-sqrt(33))/8 ,t=(9+sqrt(33))/8

Find the zeros of 15=3+36t-16t^2

The zeros of 15=3+36t-16t^2 are t=(9-sqrt(33))/8 ,t=(9+sqrt(33))/8

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15 = 3 + 36 t - 16 t^{2}

t = \frac{9 - 33}{8}, t = \frac{9 + 33}{8}

Decimal Notation

t = 0.40692 \dots, t = 1.84307 \dots

Solution steps

15 = 3 + 36 t - 16 t^{2}

Switch sides

3 + 36 t - 16 t^{2} = 15

Move

15

to the left side

- 16 t^{2} + 36 t - 12 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 36 \pm 3 6 ^{2} - 4 ( - 16 ) ( - 12 )}{2 ( - 16 )}

3 6^{2} - 4 (- 16) (- 12) = 433

t_{1, 2} = \frac{- 36 \pm 4 33}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 36 + 4 33}{2 ( - 16 )}, t_{2} = \frac{- 36 - 4 33}{2 ( - 16 )}

t = \frac{- 36 + 4 33}{2 ( - 16 )} : \frac{9 - 33}{8}

t = \frac{- 36 - 4 33}{2 ( - 16 )} : \frac{9 + 33}{8}

The solutions to the quadratic equation are:

t = \frac{9 - 33}{8}, t = \frac{9 + 33}{8}

x^{2} + 6 = 8

2 c - 4 > - 6

(\frac{2}{3}) 2

(x + 1) (x - 4) \leq 0

2 - (- 2)^{2}

What are the solutions to the equation 15=3+36t-16t^2 ?
The solutions to the equation 15=3+36t-16t^2 are t=(9-sqrt(33))/8 ,t=(9+sqrt(33))/8
Find the zeros of 15=3+36t-16t^2
The zeros of 15=3+36t-16t^2 are t=(9-sqrt(33))/8 ,t=(9+sqrt(33))/8