What are the solutions to the equation 39=113t-16t^2 ?

The solutions to the equation 39=113t-16t^2 are t=(113-sqrt(10273))/(32),t=(113+sqrt(10273))/(32)

Find the zeros of 39=113t-16t^2

The zeros of 39=113t-16t^2 are t=(113-sqrt(10273))/(32),t=(113+sqrt(10273))/(32)

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39 = 113 t - 16 t^{2}

t = \frac{113 - 10273}{32}, t = \frac{113 + 10273}{32}

Decimal Notation

t = 0.36388 \dots, t = 6.69861 \dots

Solution steps

39 = 113 t - 16 t^{2}

Switch sides

113 t - 16 t^{2} = 39

Move

39

to the left side

113 t - 16 t^{2} - 39 = 0

Write in the standard form

a x^{2} + b x + c = 0

- 16 t^{2} + 113 t - 39 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 113 \pm 11 3 ^{2} - 4 ( - 16 ) ( - 39 )}{2 ( - 16 )}

11 3^{2} - 4 (- 16) (- 39) = 10273

t_{1, 2} = \frac{- 113 \pm 10273}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 113 + 10273}{2 ( - 16 )}, t_{2} = \frac{- 113 - 10273}{2 ( - 16 )}

t = \frac{- 113 + 10273}{2 ( - 16 )} : \frac{113 - 10273}{32}

t = \frac{- 113 - 10273}{2 ( - 16 )} : \frac{113 + 10273}{32}

The solutions to the quadratic equation are:

t = \frac{113 - 10273}{32}, t = \frac{113 + 10273}{32}

x^{2} + 7 x + 3 = 3

- 4 t^{2} + 32 t = - 132

x^{2} + (4 x + 16)^{2} = (4 x + 17)^{2}

x^{2} + 2 x = 80

x^{2} + 2 x = 15

What are the solutions to the equation 39=113t-16t^2 ?
The solutions to the equation 39=113t-16t^2 are t=(113-sqrt(10273))/(32),t=(113+sqrt(10273))/(32)
Find the zeros of 39=113t-16t^2
The zeros of 39=113t-16t^2 are t=(113-sqrt(10273))/(32),t=(113+sqrt(10273))/(32)