What are the solutions to the equation 123-11t-16t^2=0 ?

The solutions to the equation 123-11t-16t^2=0 are t=-(11+sqrt(7993))/(32),t=(sqrt(7993)-11)/(32)

Find the zeros of 123-11t-16t^2=0

The zeros of 123-11t-16t^2=0 are t=-(11+sqrt(7993))/(32),t=(sqrt(7993)-11)/(32)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

123 - 11 t - 16 t^{2} = 0

t = - \frac{11 + 7993}{32}, t = \frac{7993 - 11}{32}

Decimal Notation

t = - 3.13761 \dots, t = 2.45011 \dots

Solution steps

123 - 11 t - 16 t^{2} = 0

Write in the standard form

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

- 16 t^{2} - 11 t + 123 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 ( - 16 ) \cdot 123}{2 ( - 16 )}

(- 11)^{2} - 4 (- 16) \cdot 123 = 7993

t_{1, 2} = \frac{- ( - 11 ) \pm 7993}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- ( - 11 ) + 7993}{2 ( - 16 )}, t_{2} = \frac{- ( - 11 ) - 7993}{2 ( - 16 )}

t = \frac{- ( - 11 ) + 7993}{2 ( - 16 )} : - \frac{11 + 7993}{32}

t = \frac{- ( - 11 ) - 7993}{2 ( - 16 )} : \frac{7993 - 11}{32}

The solutions to the quadratic equation are:

t = - \frac{11 + 7993}{32}, t = \frac{7993 - 11}{32}

\frac{1}{4} + \frac{2}{3}

(\frac{11}{2})^{2}

6 x^{2} - 2 x - 8 = 0

\frac{- 13.4}{- 9.35}

3 (2 x - 4) > 4 (- 2 x + 1)

What are the solutions to the equation 123-11t-16t^2=0 ?
The solutions to the equation 123-11t-16t^2=0 are t=-(11+sqrt(7993))/(32),t=(sqrt(7993)-11)/(32)
Find the zeros of 123-11t-16t^2=0
The zeros of 123-11t-16t^2=0 are t=-(11+sqrt(7993))/(32),t=(sqrt(7993)-11)/(32)