What are the solutions to the equation t^2-8t+15=0 ?

The solutions to the equation t^2-8t+15=0 are t=5,t=3

Find the zeros of t^2-8t+15=0

The zeros of t^2-8t+15=0 are t=5,t=3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

t^{2} - 8 t + 15 = 0

t = 5, t = 3

Solution steps

t^{2} - 8 t + 15 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 1 \cdot 15}{2 \cdot 1}

(- 8)^{2} - 4 \cdot 1 \cdot 15 = 2

t_{1, 2} = \frac{- ( - 8 ) \pm 2}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- ( - 8 ) + 2}{2 \cdot 1}, t_{2} = \frac{- ( - 8 ) - 2}{2 \cdot 1}

t = \frac{- ( - 8 ) + 2}{2 \cdot 1} : 5

t = \frac{- ( - 8 ) - 2}{2 \cdot 1} : 3

The solutions to the quadratic equation are:

t = 5, t = 3

80 x^{5} - 125 x

49 x^{8} - 25 x^{6}

\frac{4 π}{2}

10 x - 3 (x + 2) = 1

4 (x + 8)^{2} - 40 = - 36

What are the solutions to the equation t^2-8t+15=0 ?
The solutions to the equation t^2-8t+15=0 are t=5,t=3
Find the zeros of t^2-8t+15=0
The zeros of t^2-8t+15=0 are t=5,t=3