What are the solutions to the equation 0=x^2-4x-3 ?

The solutions to the equation 0=x^2-4x-3 are x=2+sqrt(7),x=2-sqrt(7)

Find the zeros of 0=x^2-4x-3

The zeros of 0=x^2-4x-3 are x=2+sqrt(7),x=2-sqrt(7)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

0 = x^{2} - 4 x - 3

x = 2 + 7, x = 2 - 7

Solution steps

0 = x^{2} - 4 x - 3

Switch sides

x^{2} - 4 x - 3 = 0

Solve with the quadratic formula

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

(- 4)^{2} - 4 \cdot 1 \cdot (- 3) = 27

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

Separate the solutions

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

x = \frac{- ( - 4 ) + 2 7}{2 \cdot 1} : 2 + 7

x = \frac{- ( - 4 ) - 2 7}{2 \cdot 1} : 2 - 7

The solutions to the quadratic equation are:

x = 2 + 7, x = 2 - 7

\frac{( 6 + \frac{2 y}{4} )}{11} - \frac{y}{8} = - 2

4^{2 x} - 2 (4^{x}) - 8

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

x^{4} - 13 x^{2} + 36 < 0

∣ 5 x - 1 ∣ = 2

What are the solutions to the equation 0=x^2-4x-3 ?
The solutions to the equation 0=x^2-4x-3 are x=2+sqrt(7),x=2-sqrt(7)
Find the zeros of 0=x^2-4x-3
The zeros of 0=x^2-4x-3 are x=2+sqrt(7),x=2-sqrt(7)