What are the solutions to the equation -16t^2+160t+120=0 ?

The solutions to the equation -16t^2+160t+120=0 are t=-(-10+sqrt(130))/2 ,t=(10+sqrt(130))/2

Find the zeros of -16t^2+160t+120=0

The zeros of -16t^2+160t+120=0 are t=-(-10+sqrt(130))/2 ,t=(10+sqrt(130))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

- 16 t^{2} + 160 t + 120 = 0

t = - \frac{- 10 + 130}{2}, t = \frac{10 + 130}{2}

Decimal Notation

t = - 0.70087 \dots, t = 10.70087 \dots

Solution steps

- 16 t^{2} + 160 t + 120 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 160 \pm 16 0 ^{2} - 4 ( - 16 ) \cdot 120}{2 ( - 16 )}

16 0^{2} - 4 (- 16) \cdot 120 = 16130

t_{1, 2} = \frac{- 160 \pm 16 130}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 160 + 16 130}{2 ( - 16 )}, t_{2} = \frac{- 160 - 16 130}{2 ( - 16 )}

t = \frac{- 160 + 16 130}{2 ( - 16 )} : - \frac{- 10 + 130}{2}

t = \frac{- 160 - 16 130}{2 ( - 16 )} : \frac{10 + 130}{2}

The solutions to the quadratic equation are:

t = - \frac{- 10 + 130}{2}, t = \frac{10 + 130}{2}

5 + 18 x = 35 x^{2}

x, x + y - x^{2} + 2 x y - y^{2} = 0

\frac{3}{2} = x^{2}

0 = 3 x^{2} + 12 x + 9

z, z^{2} = x^{2} + y^{2}

What are the solutions to the equation -16t^2+160t+120=0 ?
The solutions to the equation -16t^2+160t+120=0 are t=-(-10+sqrt(130))/2 ,t=(10+sqrt(130))/2
Find the zeros of -16t^2+160t+120=0
The zeros of -16t^2+160t+120=0 are t=-(-10+sqrt(130))/2 ,t=(10+sqrt(130))/2