What are the solutions to the equation x^2+7x-4=0 ?

The solutions to the equation x^2+7x-4=0 are x=(-7+sqrt(65))/2 ,x=(-7-sqrt(65))/2

Find the zeros of x^2+7x-4=0

The zeros of x^2+7x-4=0 are x=(-7+sqrt(65))/2 ,x=(-7-sqrt(65))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^{2} + 7 x - 4 = 0

x = \frac{- 7 + 65}{2}, x = \frac{- 7 - 65}{2}

Decimal Notation

x = 0.53112 \dots, x = - 7.53112 \dots

Solution steps

x^{2} + 7 x - 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 1 \cdot ( - 4 )}{2 \cdot 1}

7^{2} - 4 \cdot 1 \cdot (- 4) = 65

x_{1, 2} = \frac{- 7 \pm 65}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 7 + 65}{2 \cdot 1}, x_{2} = \frac{- 7 - 65}{2 \cdot 1}

x = \frac{- 7 + 65}{2 \cdot 1} : \frac{- 7 + 65}{2}

x = \frac{- 7 - 65}{2 \cdot 1} : \frac{- 7 - 65}{2}

The solutions to the quadratic equation are:

x = \frac{- 7 + 65}{2}, x = \frac{- 7 - 65}{2}

7 x^{2} + 3 x = 8

\frac{( x - 5 ) ( 2 x + 3 )}{( x - 1 ) ^{2}} \leq 0

64 u^{3} - v^{3}

2 m^{2} - 11 m + 15

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

What are the solutions to the equation x^2+7x-4=0 ?
The solutions to the equation x^2+7x-4=0 are x=(-7+sqrt(65))/2 ,x=(-7-sqrt(65))/2
Find the zeros of x^2+7x-4=0
The zeros of x^2+7x-4=0 are x=(-7+sqrt(65))/2 ,x=(-7-sqrt(65))/2