What are the solutions to the equation x^2-x(sqrt(6)-1)-sqrt(6)=0 ?

The solutions to the equation x^2-x(sqrt(6)-1)-sqrt(6)=0 are x=sqrt(6),x=-1

Find the zeros of x^2-x(sqrt(6)-1)-sqrt(6)=0

The zeros of x^2-x(sqrt(6)-1)-sqrt(6)=0 are x=sqrt(6),x=-1

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x^{2} - x (6 - 1) - 6 = 0

x = 6, x = - 1

Decimal Notation

x = 2.44948 \dots, x = - 1

Solution steps

x^{2} - x (6 - 1) - 6 = 0

Expand

x^{2} - x (6 - 1) - 6 : x^{2} - 6 x + x - 6

x^{2} - 6 x + x - 6 = 0

Write in the standard form

a x^{2} + b x + c = 0

x^{2} + (- 6 + 1) x - 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 6 + 1 ) \pm ( - 6 + 1 ) ^{2} - 4 \cdot 1 \cdot ( - 6 )}{2 \cdot 1}

(- 6 + 1)^{2} - 4 \cdot 1 \cdot (- 6) = 6 + 1

x_{1, 2} = \frac{- ( - 6 + 1 ) \pm ( 6 + 1 )}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 6 + 1 ) + 6 + 1}{2 \cdot 1}, x_{2} = \frac{- ( - 6 + 1 ) - ( 6 + 1 )}{2 \cdot 1}

x = \frac{- ( - 6 + 1 ) + 6 + 1}{2 \cdot 1} : 6

x = \frac{- ( - 6 + 1 ) - ( 6 + 1 )}{2 \cdot 1} : - 1

The solutions to the quadratic equation are:

x = 6, x = - 1

x, 2 x^{2} + p^{2} = (q - x)^{2}

a \cdot x^{2} - 8 x + 4 = 0

x, x^{2} y^{2} + x^{2} + y^{2} = 3736

x^{2} = 4 \frac{x}{5} + \frac{1}{5}

(m - 1) \cdot x^{2} - 2 (m - 1) x + 1 = 0

What are the solutions to the equation x^2-x(sqrt(6)-1)-sqrt(6)=0 ?
The solutions to the equation x^2-x(sqrt(6)-1)-sqrt(6)=0 are x=sqrt(6),x=-1
Find the zeros of x^2-x(sqrt(6)-1)-sqrt(6)=0
The zeros of x^2-x(sqrt(6)-1)-sqrt(6)=0 are x=sqrt(6),x=-1