What are the solutions to the equation 2x^2+3x-7=0 ?

The solutions to the equation 2x^2+3x-7=0 are x=(-3+sqrt(65))/4 ,x=(-3-sqrt(65))/4

Find the zeros of 2x^2+3x-7=0

The zeros of 2x^2+3x-7=0 are x=(-3+sqrt(65))/4 ,x=(-3-sqrt(65))/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 - 7 = 0

x = \frac{- 3 + 65}{4}, x = \frac{- 3 - 65}{4}

Decimal Notation

x = 1.26556 \dots, x = - 2.76556 \dots

Solution steps

2 x^{2} + 3 x - 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 2 ( - 7 )}{2 \cdot 2}

3^{2} - 4 \cdot 2 (- 7) = 65

x_{1, 2} = \frac{- 3 \pm 65}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 3 + 65}{2 \cdot 2}, x_{2} = \frac{- 3 - 65}{2 \cdot 2}

x = \frac{- 3 + 65}{2 \cdot 2} : \frac{- 3 + 65}{4}

x = \frac{- 3 - 65}{2 \cdot 2} : \frac{- 3 - 65}{4}

The solutions to the quadratic equation are:

x = \frac{- 3 + 65}{4}, x = \frac{- 3 - 65}{4}

- 4 x = 16

3 x^{2} + 14 = 12

\frac{3}{7} + \frac{13}{4} \cdot \frac{1}{3}

lo g_{25} (49)

\frac{( 2 x - 3 )}{6} - 2 (\frac{3}{2} x - 1) - \frac{3 - 2 x}{3} = 0

What are the solutions to the equation 2x^2+3x-7=0 ?
The solutions to the equation 2x^2+3x-7=0 are x=(-3+sqrt(65))/4 ,x=(-3-sqrt(65))/4
Find the zeros of 2x^2+3x-7=0
The zeros of 2x^2+3x-7=0 are x=(-3+sqrt(65))/4 ,x=(-3-sqrt(65))/4