What are the solutions to the equation (2-t)*(3-t)-2=0 ?

The solutions to the equation (2-t)*(3-t)-2=0 are t=4,t=1

Find the zeros of (2-t)*(3-t)-2=0

The zeros of (2-t)*(3-t)-2=0 are t=4,t=1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(2 - t) \cdot (3 - t) - 2 = 0

t = 4, t = 1

Solution steps

(2 - t) (3 - t) - 2 = 0

Expand

(2 - t) (3 - t) - 2 : t^{2} - 5 t + 4

t^{2} - 5 t + 4 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 1 \cdot 4}{2 \cdot 1}

(- 5)^{2} - 4 \cdot 1 \cdot 4 = 3

t_{1, 2} = \frac{- ( - 5 ) \pm 3}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- ( - 5 ) + 3}{2 \cdot 1}, t_{2} = \frac{- ( - 5 ) - 3}{2 \cdot 1}

t = \frac{- ( - 5 ) + 3}{2 \cdot 1} : 4

t = \frac{- ( - 5 ) - 3}{2 \cdot 1} : 1

The solutions to the quadratic equation are:

t = 4, t = 1

- 0.5 (t)^{2} + t \cdot 3.2 + 22 = 24

(2 x)^{2} - 60 = 4

x, k x^{2} + 2 x - k = k x - 2

(z + 6)^{2} = - 36

x^{2} - b x + 16 = 0

What are the solutions to the equation (2-t)*(3-t)-2=0 ?
The solutions to the equation (2-t)*(3-t)-2=0 are t=4,t=1
Find the zeros of (2-t)*(3-t)-2=0
The zeros of (2-t)*(3-t)-2=0 are t=4,t=1