What are the solutions to the equation t^2-9t+10=0 ?

The solutions to the equation t^2-9t+10=0 are t=(9+sqrt(41))/2 ,t=(9-sqrt(41))/2

Find the zeros of t^2-9t+10=0

The zeros of t^2-9t+10=0 are t=(9+sqrt(41))/2 ,t=(9-sqrt(41))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

t^{2} - 9 t + 10 = 0

t = \frac{9 + 41}{2}, t = \frac{9 - 41}{2}

Decimal Notation

t = 7.70156 \dots, t = 1.29843 \dots

Solution steps

t^{2} - 9 t + 10 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 1 \cdot 10}{2 \cdot 1}

(- 9)^{2} - 4 \cdot 1 \cdot 10 = 41

t_{1, 2} = \frac{- ( - 9 ) \pm 41}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- ( - 9 ) + 41}{2 \cdot 1}, t_{2} = \frac{- ( - 9 ) - 41}{2 \cdot 1}

t = \frac{- ( - 9 ) + 41}{2 \cdot 1} : \frac{9 + 41}{2}

t = \frac{- ( - 9 ) - 41}{2 \cdot 1} : \frac{9 - 41}{2}

The solutions to the quadratic equation are:

t = \frac{9 + 41}{2}, t = \frac{9 - 41}{2}

x^{2} - 3 x \leq 18

(- 2 x)^{- 2}

t = - \frac{ln ( 2 )}{ln ( \frac{298}{500} )} \cdot 8

a^{2} - 3 a

(- 30 i) + (52 - 30 i)

What are the solutions to the equation t^2-9t+10=0 ?
The solutions to the equation t^2-9t+10=0 are t=(9+sqrt(41))/2 ,t=(9-sqrt(41))/2
Find the zeros of t^2-9t+10=0
The zeros of t^2-9t+10=0 are t=(9+sqrt(41))/2 ,t=(9-sqrt(41))/2