What are the solutions to the equation 2x^2+10x-4=0 ?

The solutions to the equation 2x^2+10x-4=0 are x=(-5+sqrt(33))/2 ,x=-(5+sqrt(33))/2

Find the zeros of 2x^2+10x-4=0

The zeros of 2x^2+10x-4=0 are x=(-5+sqrt(33))/2 ,x=-(5+sqrt(33))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2 x^{2} + 10 x - 4 = 0

x = \frac{- 5 + 33}{2}, x = - \frac{5 + 33}{2}

Decimal Notation

x = 0.37228 \dots, x = - 5.37228 \dots

Solution steps

2 x^{2} + 10 x - 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 2 ( - 4 )}{2 \cdot 2}

1 0^{2} - 4 \cdot 2 (- 4) = 233

x_{1, 2} = \frac{- 10 \pm 2 33}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 10 + 2 33}{2 \cdot 2}, x_{2} = \frac{- 10 - 2 33}{2 \cdot 2}

x = \frac{- 10 + 2 33}{2 \cdot 2} : \frac{- 5 + 33}{2}

x = \frac{- 10 - 2 33}{2 \cdot 2} : - \frac{5 + 33}{2}

The solutions to the quadratic equation are:

x = \frac{- 5 + 33}{2}, x = - \frac{5 + 33}{2}

\frac{- 8 x ^{4} y}{24 x ^{9} y ^{8}}

e^{u} - (u - 1) \cdot e^{u}

m^{2} n^{2} + m n^{2} - 2 m^{2} n

(\frac{1}{N} + \frac{0.0004}{1 - 0.0004 N}) d N = d t

x^{2} - 10 x + 24 > 0

What are the solutions to the equation 2x^2+10x-4=0 ?
The solutions to the equation 2x^2+10x-4=0 are x=(-5+sqrt(33))/2 ,x=-(5+sqrt(33))/2
Find the zeros of 2x^2+10x-4=0
The zeros of 2x^2+10x-4=0 are x=(-5+sqrt(33))/2 ,x=-(5+sqrt(33))/2