What are the solutions to the equation quadraticformula 2x^2-9x+5=0 ?

The solutions to the equation quadraticformula 2x^2-9x+5=0 are x=(9+sqrt(41))/4 ,x=(9-sqrt(41))/4

Find the zeros of quadraticformula 2x^2-9x+5=0

The zeros of quadraticformula 2x^2-9x+5=0 are x=(9+sqrt(41))/4 ,x=(9-sqrt(41))/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2 x^{2} - 9 x + 5 = 0

x = \frac{9 + 41}{4}, x = \frac{9 - 41}{4}

Decimal Notation

x = 3.85078 \dots, x = 0.64921 \dots

Solution steps

2 x^{2} - 9 x + 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 2 \cdot 5}{2 \cdot 2}

(- 9)^{2} - 4 \cdot 2 \cdot 5 = 41

x_{1, 2} = \frac{- ( - 9 ) \pm 41}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 9 ) + 41}{2 \cdot 2}, x_{2} = \frac{- ( - 9 ) - 41}{2 \cdot 2}

x = \frac{- ( - 9 ) + 41}{2 \cdot 2} : \frac{9 + 41}{4}

x = \frac{- ( - 9 ) - 41}{2 \cdot 2} : \frac{9 - 41}{4}

The solutions to the quadratic equation are:

x = \frac{9 + 41}{4}, x = \frac{9 - 41}{4}

6 x^{2} - 7 x = - 2

w^{2} = - 121

v, F = \frac{j M v ^{2}}{r}

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

(x - 4)^{2} + (x + 4)^{2} = 40

What are the solutions to the equation quadraticformula 2x^2-9x+5=0 ?
The solutions to the equation quadraticformula 2x^2-9x+5=0 are x=(9+sqrt(41))/4 ,x=(9-sqrt(41))/4
Find the zeros of quadraticformula 2x^2-9x+5=0
The zeros of quadraticformula 2x^2-9x+5=0 are x=(9+sqrt(41))/4 ,x=(9-sqrt(41))/4