What are the solutions to the equation (t-2)(t+5)=t-2(t+5) ?

The solutions to the equation (t-2)(t+5)=t-2(t+5) are t=0,t=-4

Find the zeros of (t-2)(t+5)=t-2(t+5)

The zeros of (t-2)(t+5)=t-2(t+5) are t=0,t=-4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(t - 2) (t + 5) = t - 2 (t + 5)

t = 0, t = - 4

Solution steps

(t - 2) (t + 5) = t - 2 (t + 5)

Expand

(t - 2) (t + 5) : t^{2} + 3 t - 10

Expand

t - 2 (t + 5) : - t - 10

t^{2} + 3 t - 10 = - t - 10

Move

10

to the left side

t^{2} + 3 t = - t

Move

t

to the left side

t^{2} + 4 t = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

4^{2} - 4 \cdot 1 \cdot 0 = 4

t_{1, 2} = \frac{- 4 \pm 4}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- 4 + 4}{2 \cdot 1}, t_{2} = \frac{- 4 - 4}{2 \cdot 1}

t = \frac{- 4 + 4}{2 \cdot 1} : 0

t = \frac{- 4 - 4}{2 \cdot 1} : - 4

The solutions to the quadratic equation are:

t = 0, t = - 4

x^{2} - 2 x = 63

x - 6 = - 10

5 x + 6 = 3 x + 7

(x^{\frac{1}{6}} y^{\frac{1}{3}})^{- 18}

- 16 - 4 u = - 16

What are the solutions to the equation (t-2)(t+5)=t-2(t+5) ?
The solutions to the equation (t-2)(t+5)=t-2(t+5) are t=0,t=-4
Find the zeros of (t-2)(t+5)=t-2(t+5)
The zeros of (t-2)(t+5)=t-2(t+5) are t=0,t=-4