What are the solutions to the equation 2x^2+3x=4 ?

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

Find the zeros of 2x^2+3x=4

The zeros of 2x^2+3x=4 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 = 4

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

Decimal Notation

x = 0.85078 \dots, x = - 2.35078 \dots

Solution steps

2 x^{2} + 3 x = 4

Move

4

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}

(\frac{x ^{2}}{8})^{2}

x^{2} + 5^{2}

\frac{1}{8}

\frac{3 x}{10} = \frac{9}{4}

\frac{( 6 4 ^{\frac{5}{9}} \cdot 6 4 ^{\frac{2}{9}} )}{4 ^{\frac{3}{4}}}

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