What are the solutions to the equation x^2+3x-5=0 ?

The solutions to the equation x^2+3x-5=0 are x=(-3+sqrt(29))/2 ,x=(-3-sqrt(29))/2

Find the zeros of x^2+3x-5=0

The zeros of x^2+3x-5=0 are x=(-3+sqrt(29))/2 ,x=(-3-sqrt(29))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^{2} + 3 x - 5 = 0

x = \frac{- 3 + 29}{2}, x = \frac{- 3 - 29}{2}

Decimal Notation

x = 1.19258 \dots, x = - 4.19258 \dots

Solution steps

x^{2} + 3 x - 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot ( - 5 )}{2 \cdot 1}

3^{2} - 4 \cdot 1 \cdot (- 5) = 29

x_{1, 2} = \frac{- 3 \pm 29}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 3 + 29}{2 \cdot 1}, x_{2} = \frac{- 3 - 29}{2 \cdot 1}

x = \frac{- 3 + 29}{2 \cdot 1} : \frac{- 3 + 29}{2}

x = \frac{- 3 - 29}{2 \cdot 1} : \frac{- 3 - 29}{2}

The solutions to the quadratic equation are:

x = \frac{- 3 + 29}{2}, x = \frac{- 3 - 29}{2}

x^{2} + x - 56 < 0

5 x^{3} + 30 x^{2} + 40 x

16.8 = \frac{x}{3.09}

n^{2} + 14 n + 13

\frac{( 2 t + 5 ) ( 3 t - 1 )}{3} + \frac{t ^{2} + 5}{2} = \frac{7 t - 5}{6}

What are the solutions to the equation x^2+3x-5=0 ?
The solutions to the equation x^2+3x-5=0 are x=(-3+sqrt(29))/2 ,x=(-3-sqrt(29))/2
Find the zeros of x^2+3x-5=0
The zeros of x^2+3x-5=0 are x=(-3+sqrt(29))/2 ,x=(-3-sqrt(29))/2