What are the solutions to the equation 16-x^2=x(8+2) ?

The solutions to the equation 16-x^2=x(8+2) are x=-5-sqrt(41),x=sqrt(41)-5

Find the zeros of 16-x^2=x(8+2)

The zeros of 16-x^2=x(8+2) are x=-5-sqrt(41),x=sqrt(41)-5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

16 - x^{2} = x (8 + 2)

x = - 5 - 41, x = 41 - 5

Decimal Notation

x = - 11.40312 \dots, x = 1.40312 \dots

Solution steps

16 - x^{2} = x (8 + 2)

Expand

x (8 + 2) : 10 x

16 - x^{2} = 10 x

Move

10 x

to the left side

16 - x^{2} - 10 x = 0

Write in the standard form

a x^{2} + b x + c = 0

- x^{2} - 10 x + 16 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 ( - 1 ) \cdot 16}{2 ( - 1 )}

(- 10)^{2} - 4 (- 1) \cdot 16 = 241

x_{1, 2} = \frac{- ( - 10 ) \pm 2 41}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- ( - 10 ) + 2 41}{2 ( - 1 )}, x_{2} = \frac{- ( - 10 ) - 2 41}{2 ( - 1 )}

x = \frac{- ( - 10 ) + 2 41}{2 ( - 1 )} : - 5 - 41

x = \frac{- ( - 10 ) - 2 41}{2 ( - 1 )} : 41 - 5

The solutions to the quadratic equation are:

x = - 5 - 41, x = 41 - 5

\frac{1}{3} x^{2} + \frac{123}{4} x - \frac{1}{6} = 0

x^{2} - 20 x + 116 = 0

r^{2} - 123 r + 80 = 0

0 = x^{2} + 12 x + 20

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

What are the solutions to the equation 16-x^2=x(8+2) ?
The solutions to the equation 16-x^2=x(8+2) are x=-5-sqrt(41),x=sqrt(41)-5
Find the zeros of 16-x^2=x(8+2)
The zeros of 16-x^2=x(8+2) are x=-5-sqrt(41),x=sqrt(41)-5