What are the solutions to the equation 95=183t-16t^2 ?

The solutions to the equation 95=183t-16t^2 are t=(183-sqrt(27409))/(32),t=(183+sqrt(27409))/(32)

Find the zeros of 95=183t-16t^2

The zeros of 95=183t-16t^2 are t=(183-sqrt(27409))/(32),t=(183+sqrt(27409))/(32)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

95 = 183 t - 16 t^{2}

t = \frac{183 - 27409}{32}, t = \frac{183 + 27409}{32}

Decimal Notation

t = 0.54510 \dots, t = 10.89239 \dots

Solution steps

95 = 183 t - 16 t^{2}

Switch sides

183 t - 16 t^{2} = 95

Move

95

to the left side

183 t - 16 t^{2} - 95 = 0

Write in the standard form

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

- 16 t^{2} + 183 t - 95 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 183 \pm 18 3 ^{2} - 4 ( - 16 ) ( - 95 )}{2 ( - 16 )}

18 3^{2} - 4 (- 16) (- 95) = 27409

t_{1, 2} = \frac{- 183 \pm 27409}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 183 + 27409}{2 ( - 16 )}, t_{2} = \frac{- 183 - 27409}{2 ( - 16 )}

t = \frac{- 183 + 27409}{2 ( - 16 )} : \frac{183 - 27409}{32}

t = \frac{- 183 - 27409}{2 ( - 16 )} : \frac{183 + 27409}{32}

The solutions to the quadratic equation are:

t = \frac{183 - 27409}{32}, t = \frac{183 + 27409}{32}

x^{2} - 4 x - 144 = 0

∣ 2 x + 3 ∣ + 3 x = 1

12 x + 36 = 144

(\frac{49}{36})^{\frac{3}{2}}

3 (x + 7) \geq 2 x + 6

What are the solutions to the equation 95=183t-16t^2 ?
The solutions to the equation 95=183t-16t^2 are t=(183-sqrt(27409))/(32),t=(183+sqrt(27409))/(32)
Find the zeros of 95=183t-16t^2
The zeros of 95=183t-16t^2 are t=(183-sqrt(27409))/(32),t=(183+sqrt(27409))/(32)