What are the solutions to the equation 6+(5x)(2x)-8+2x=2x(x+1) ?

The solutions to the equation 6+(5x)(2x)-8+2x=2x(x+1) are x= 1/2 ,x=-1/2

Find the zeros of 6+(5x)(2x)-8+2x=2x(x+1)

The zeros of 6+(5x)(2x)-8+2x=2x(x+1) are x= 1/2 ,x=-1/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

6 + (5 x) (2 x) - 8 + 2 x = 2 x (x + 1)

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

Decimal Notation

x = 0.5, x = - 0.5

Solution steps

6 + (5 x) (2 x) - 8 + 2 x = 2 x (x + 1)

Expand

6 + (5 x) (2 x) - 8 + 2 x : 10 x^{2} + 2 x - 2

Expand

2 x (x + 1) : 2 x^{2} + 2 x

10 x^{2} + 2 x - 2 = 2 x^{2} + 2 x

Move

2

to the right side

10 x^{2} + 2 x = 2 x^{2} + 2 x + 2

Move

2 x

to the left side

10 x^{2} = 2 x^{2} + 2

Move

2 x^{2}

to the left side

8 x^{2} = 2

Divide both sides by

8

x^{2} = \frac{1}{4}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

x = \frac{1}{4}, x = - \frac{1}{4}

\frac{1}{4} = \frac{1}{2}

- \frac{1}{4} = - \frac{1}{2}

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

y, x^{3} - 2 y^{2} - 9 = x

\frac{9}{16} - \frac{x ^{2}}{81} = 0

0.4 x^{2} + 1.5 x - 1.3 = 0

s^{2} + s + 5

(- \frac{11}{4})^{2} + y^{2} = 1

What are the solutions to the equation 6+(5x)(2x)-8+2x=2x(x+1) ?
The solutions to the equation 6+(5x)(2x)-8+2x=2x(x+1) are x= 1/2 ,x=-1/2
Find the zeros of 6+(5x)(2x)-8+2x=2x(x+1)
The zeros of 6+(5x)(2x)-8+2x=2x(x+1) are x= 1/2 ,x=-1/2