What are the solutions to the equation x^2+5x-22=6x+2 ?

The solutions to the equation x^2+5x-22=6x+2 are x=(1+sqrt(97))/2 ,x=(1-sqrt(97))/2

Find the zeros of x^2+5x-22=6x+2

The zeros of x^2+5x-22=6x+2 are x=(1+sqrt(97))/2 ,x=(1-sqrt(97))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^{2} + 5 x - 22 = 6 x + 2

x = \frac{1 + 97}{2}, x = \frac{1 - 97}{2}

Decimal Notation

x = 5.42442 \dots, x = - 4.42442 \dots

Solution steps

x^{2} + 5 x - 22 = 6 x + 2

Move

2

to the left side

x^{2} + 5 x - 24 = 6 x

Move

6 x

to the left side

x^{2} - x - 24 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot ( - 24 )}{2 \cdot 1}

(- 1)^{2} - 4 \cdot 1 \cdot (- 24) = 97

x_{1, 2} = \frac{- ( - 1 ) \pm 97}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 97}{2 \cdot 1}, x_{2} = \frac{- ( - 1 ) - 97}{2 \cdot 1}

x = \frac{- ( - 1 ) + 97}{2 \cdot 1} : \frac{1 + 97}{2}

x = \frac{- ( - 1 ) - 97}{2 \cdot 1} : \frac{1 - 97}{2}

The solutions to the quadratic equation are:

x = \frac{1 + 97}{2}, x = \frac{1 - 97}{2}

x, x^{2} + y^{2} = x^{2} y^{2}

x, 4 x^{2} + 4 y^{2} + 12 x - 24 y + 45 = 0

3 + x^{2} = 0

x, A x^{2} + C y^{2} + D x + E y + F = 0

\frac{( ( 2 x - 1 ) ^{2} )}{3} - \frac{( x ^{2} - 5 )}{8} = \frac{7}{2}

What are the solutions to the equation x^2+5x-22=6x+2 ?
The solutions to the equation x^2+5x-22=6x+2 are x=(1+sqrt(97))/2 ,x=(1-sqrt(97))/2
Find the zeros of x^2+5x-22=6x+2
The zeros of x^2+5x-22=6x+2 are x=(1+sqrt(97))/2 ,x=(1-sqrt(97))/2