What are the solutions to the equation x^2-4x-7=1 ?

The solutions to the equation x^2-4x-7=1 are x=2(1+sqrt(3)),x=2(1-sqrt(3))

Find the zeros of x^2-4x-7=1

The zeros of x^2-4x-7=1 are x=2(1+sqrt(3)),x=2(1-sqrt(3))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^{2} - 4 x - 7 = 1

x = 2 (1 + 3), x = 2 (1 - 3)

Decimal Notation

x = 5.46410 \dots, x = - 1.46410 \dots

Solution steps

x^{2} - 4 x - 7 = 1

Move

1

to the left side

x^{2} - 4 x - 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot ( - 8 )}{2 \cdot 1}

(- 4)^{2} - 4 \cdot 1 \cdot (- 8) = 43

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

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 4 3}{2 \cdot 1}, x_{2} = \frac{- ( - 4 ) - 4 3}{2 \cdot 1}

x = \frac{- ( - 4 ) + 4 3}{2 \cdot 1} : 2 (1 + 3)

x = \frac{- ( - 4 ) - 4 3}{2 \cdot 1} : 2 (1 - 3)

The solutions to the quadratic equation are:

x = 2 (1 + 3), x = 2 (1 - 3)

(x - 3)^{2} - 20 = 0

2 x^{3} + 7 x^{2} + 6 x

2 n^{2} + 9 n + 9

- \frac{19}{2} (\frac{1}{10 0 ^{2}}) + \frac{19}{2}

\frac{2}{5} \cdot \frac{6}{7}

What are the solutions to the equation x^2-4x-7=1 ?
The solutions to the equation x^2-4x-7=1 are x=2(1+sqrt(3)),x=2(1-sqrt(3))
Find the zeros of x^2-4x-7=1
The zeros of x^2-4x-7=1 are x=2(1+sqrt(3)),x=2(1-sqrt(3))