What are the solutions to the equation solvefor x,2x^2+5x-4=0 ?

The solutions to the equation solvefor x,2x^2+5x-4=0 are x=(-5+sqrt(57))/4 ,x=(-5-sqrt(57))/4

Find the zeros of solvefor x,2x^2+5x-4=0

The zeros of solvefor x,2x^2+5x-4=0 are x=(-5+sqrt(57))/4 ,x=(-5-sqrt(57))/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x, 2 x^{2} + 5 x - 4 = 0

x = \frac{- 5 + 57}{4}, x = \frac{- 5 - 57}{4}

Decimal Notation

x = 0.63745 \dots, x = - 3.13745 \dots

Solution steps

2 x^{2} + 5 x - 4 = 0

Solve with the quadratic formula

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

5^{2} - 4 \cdot 2 (- 4) = 57

x_{1, 2} = \frac{- 5 \pm 57}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 5 + 57}{2 \cdot 2}, x_{2} = \frac{- 5 - 57}{2 \cdot 2}

x = \frac{- 5 + 57}{2 \cdot 2} : \frac{- 5 + 57}{4}

x = \frac{- 5 - 57}{2 \cdot 2} : \frac{- 5 - 57}{4}

The solutions to the quadratic equation are:

x = \frac{- 5 + 57}{4}, x = \frac{- 5 - 57}{4}

3 x^{2} - 1 = 11

y^{2} + 9 y - 10 = 0

250 = 2 x^{2} + \frac{x ^{2}}{2}

x, 3 x^{2} + x^{2} y = C

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

What are the solutions to the equation solvefor x,2x^2+5x-4=0 ?
The solutions to the equation solvefor x,2x^2+5x-4=0 are x=(-5+sqrt(57))/4 ,x=(-5-sqrt(57))/4
Find the zeros of solvefor x,2x^2+5x-4=0
The zeros of solvefor x,2x^2+5x-4=0 are x=(-5+sqrt(57))/4 ,x=(-5-sqrt(57))/4