What are the solutions to the equation 2x^2+3x-1=3 ?

The solutions to the equation 2x^2+3x-1=3 are x=(-3+sqrt(41))/4 ,x=(-3-sqrt(41))/4

Find the zeros of 2x^2+3x-1=3

The zeros of 2x^2+3x-1=3 are x=(-3+sqrt(41))/4 ,x=(-3-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} + 3 x - 1 = 3

x = \frac{- 3 + 41}{4}, x = \frac{- 3 - 41}{4}

Solution steps

2 x^{2} + 3 x - 1 = 3

Move

3

to the left side

2 x^{2} + 3 x - 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 2 ( - 4 )}{2 \cdot 2}

3^{2} - 4 \cdot 2 (- 4) = 41

x_{1, 2} = \frac{- 3 \pm 41}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 3 + 41}{2 \cdot 2}, x_{2} = \frac{- 3 - 41}{2 \cdot 2}

x = \frac{- 3 + 41}{2 \cdot 2} : \frac{- 3 + 41}{4}

x = \frac{- 3 - 41}{2 \cdot 2} : \frac{- 3 - 41}{4}

The solutions to the quadratic equation are:

x = \frac{- 3 + 41}{4}, x = \frac{- 3 - 41}{4}

(2 x + 1) (x + 7) = (x + 5) (3 x - 1)

\frac{2}{3} x^{2} + x + 3 = 0

(x - 8) (x - 10) = - 1

- 15 - 17 x + 4 x^{2} = 0

- x^{2} - 4 x + 10

What are the solutions to the equation 2x^2+3x-1=3 ?
The solutions to the equation 2x^2+3x-1=3 are x=(-3+sqrt(41))/4 ,x=(-3-sqrt(41))/4
Find the zeros of 2x^2+3x-1=3
The zeros of 2x^2+3x-1=3 are x=(-3+sqrt(41))/4 ,x=(-3-sqrt(41))/4