What are the solutions to the equation 2x^2+10x-5=0 ?

The solutions to the equation 2x^2+10x-5=0 are x=(-5+sqrt(35))/2 ,x=-(5+sqrt(35))/2

Find the zeros of 2x^2+10x-5=0

The zeros of 2x^2+10x-5=0 are x=(-5+sqrt(35))/2 ,x=-(5+sqrt(35))/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 - 5 = 0

x = \frac{- 5 + 35}{2}, x = - \frac{5 + 35}{2}

Decimal Notation

x = 0.45803 \dots, x = - 5.45803 \dots

Solution steps

2 x^{2} + 10 x - 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 2 ( - 5 )}{2 \cdot 2}

1 0^{2} - 4 \cdot 2 (- 5) = 235

x_{1, 2} = \frac{- 10 \pm 2 35}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 10 + 2 35}{2 \cdot 2}, x_{2} = \frac{- 10 - 2 35}{2 \cdot 2}

x = \frac{- 10 + 2 35}{2 \cdot 2} : \frac{- 5 + 35}{2}

x = \frac{- 10 - 2 35}{2 \cdot 2} : - \frac{5 + 35}{2}

The solutions to the quadratic equation are:

x = \frac{- 5 + 35}{2}, x = - \frac{5 + 35}{2}

x^{2} - x

5 (x - 3) + 6 = 5 x - 9

\frac{15 x ^{2} y - 45 x y}{15 y - 5 x y}

(- \frac{1}{3})^{4}

What are the solutions to the equation 2x^2+10x-5=0 ?
The solutions to the equation 2x^2+10x-5=0 are x=(-5+sqrt(35))/2 ,x=-(5+sqrt(35))/2
Find the zeros of 2x^2+10x-5=0
The zeros of 2x^2+10x-5=0 are x=(-5+sqrt(35))/2 ,x=-(5+sqrt(35))/2