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

The solutions to the equation x^2-5x=2 are x=(5+sqrt(33))/2 ,x=(5-sqrt(33))/2

Find the zeros of x^2-5x=2

The zeros of x^2-5x=2 are x=(5+sqrt(33))/2 ,x=(5-sqrt(33))/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 = 2

x = \frac{5 + 33}{2}, x = \frac{5 - 33}{2}

Decimal Notation

x = 5.37228 \dots, x = - 0.37228 \dots

Solution steps

x^{2} - 5 x = 2

Move

2

to the left side

x^{2} - 5 x - 2 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 1 \cdot ( - 2 )}{2 \cdot 1}

(- 5)^{2} - 4 \cdot 1 \cdot (- 2) = 33

x_{1, 2} = \frac{- ( - 5 ) \pm 33}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 33}{2 \cdot 1}, x_{2} = \frac{- ( - 5 ) - 33}{2 \cdot 1}

x = \frac{- ( - 5 ) + 33}{2 \cdot 1} : \frac{5 + 33}{2}

x = \frac{- ( - 5 ) - 33}{2 \cdot 1} : \frac{5 - 33}{2}

The solutions to the quadratic equation are:

x = \frac{5 + 33}{2}, x = \frac{5 - 33}{2}

10 - 10 ∣ - 8 k + 4 ∣ = 10

\frac{10 ( x + y ) ^{3}}{4 ( x + y ) ^{- 2}}

(7 y)^{2}

x^{\frac{3}{7}} \cdot x^{\frac{3}{14}}

What are the solutions to the equation x^2-5x=2 ?
The solutions to the equation x^2-5x=2 are x=(5+sqrt(33))/2 ,x=(5-sqrt(33))/2
Find the zeros of x^2-5x=2
The zeros of x^2-5x=2 are x=(5+sqrt(33))/2 ,x=(5-sqrt(33))/2