What are the solutions to the equation 2x^2-1=11x ?

The solutions to the equation 2x^2-1=11x are x=(11+sqrt(129))/4 ,x=(11-sqrt(129))/4

Find the zeros of 2x^2-1=11x

The zeros of 2x^2-1=11x are x=(11+sqrt(129))/4 ,x=(11-sqrt(129))/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2 x^{2} - 1 = 11 x

x = \frac{11 + 129}{4}, x = \frac{11 - 129}{4}

Decimal Notation

x = 5.58945 \dots, x = - 0.08945 \dots

Solution steps

2 x^{2} - 1 = 11 x

Move

11 x

to the left side

2 x^{2} - 1 - 11 x = 0

Write in the standard form

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

2 x^{2} - 11 x - 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 \cdot 2 ( - 1 )}{2 \cdot 2}

(- 11)^{2} - 4 \cdot 2 (- 1) = 129

x_{1, 2} = \frac{- ( - 11 ) \pm 129}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 11 ) + 129}{2 \cdot 2}, x_{2} = \frac{- ( - 11 ) - 129}{2 \cdot 2}

x = \frac{- ( - 11 ) + 129}{2 \cdot 2} : \frac{11 + 129}{4}

x = \frac{- ( - 11 ) - 129}{2 \cdot 2} : \frac{11 - 129}{4}

The solutions to the quadratic equation are:

x = \frac{11 + 129}{4}, x = \frac{11 - 129}{4}

6 4^{\frac{- 1}{6}}

y^{3} - 8 x^{3}

45 x - 45 = - 15 x + 75

∣ 3 - x ∣ = ∣ 3 x + 11 ∣

\frac{- 81}{3}

What are the solutions to the equation 2x^2-1=11x ?
The solutions to the equation 2x^2-1=11x are x=(11+sqrt(129))/4 ,x=(11-sqrt(129))/4
Find the zeros of 2x^2-1=11x
The zeros of 2x^2-1=11x are x=(11+sqrt(129))/4 ,x=(11-sqrt(129))/4