What are the solutions to the equation (x-5)(2x+1)=0 ?

The solutions to the equation (x-5)(2x+1)=0 are x=5,x=-1/2

Find the zeros of (x-5)(2x+1)=0

The zeros of (x-5)(2x+1)=0 are x=5,x=-1/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x - 5) (2 x + 1) = 0

x = 5, x = - \frac{1}{2}

Decimal Notation

x = 5, x = - 0.5

Solution steps

(x - 5) (2 x + 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 5 = 0 or 2 x + 1 = 0

Solve

x - 5 = 0 : x = 5

Solve

2 x + 1 = 0 : x = - \frac{1}{2}

The solutions to the quadratic equation are:

x = 5, x = - \frac{1}{2}

128 = \frac{2}{1125} q^{2} + 40

2 x^{1} - 3 x^{2} = - 2, 4 x^{1} - 12 x^{2} = 4

x^{2} + 3 x - 9 = 0

5 x^{2} + 3 = 128

p^{2} = 6 p - 4

What are the solutions to the equation (x-5)(2x+1)=0 ?
The solutions to the equation (x-5)(2x+1)=0 are x=5,x=-1/2
Find the zeros of (x-5)(2x+1)=0
The zeros of (x-5)(2x+1)=0 are x=5,x=-1/2