What are the solutions to the equation x^2-sqrt(7)x-6=0 ?

The solutions to the equation x^2-sqrt(7)x-6=0 are x=(sqrt(7)+sqrt(31))/2 ,x=(sqrt(7)-sqrt(31))/2

Find the zeros of x^2-sqrt(7)x-6=0

The zeros of x^2-sqrt(7)x-6=0 are x=(sqrt(7)+sqrt(31))/2 ,x=(sqrt(7)-sqrt(31))/2

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x^{2} - 7 x - 6 = 0

x = \frac{7 + 31}{2}, x = \frac{7 - 31}{2}

Decimal Notation

x = 4.10675 \dots, x = - 1.46100 \dots

Solution steps

x^{2} - 7 x - 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 1 \cdot ( - 6 )}{2 \cdot 1}

(- 7)^{2} - 4 \cdot 1 \cdot (- 6) = 31

x_{1, 2} = \frac{- ( - 7 ) \pm 31}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 7 ) + 31}{2 \cdot 1}, x_{2} = \frac{- ( - 7 ) - 31}{2 \cdot 1}

x = \frac{- ( - 7 ) + 31}{2 \cdot 1} : \frac{7 + 31}{2}

x = \frac{- ( - 7 ) - 31}{2 \cdot 1} : \frac{7 - 31}{2}

The solutions to the quadratic equation are:

x = \frac{7 + 31}{2}, x = \frac{7 - 31}{2}

lo g_{2} (4)

- 2112 + 528

5.6 j - 0.12 = 4 + 1.1 j

- \frac{1}{2} x + \frac{11}{2} = 2 x + 3

\frac{x + 1}{x - 2} > 2

What are the solutions to the equation x^2-sqrt(7)x-6=0 ?
The solutions to the equation x^2-sqrt(7)x-6=0 are x=(sqrt(7)+sqrt(31))/2 ,x=(sqrt(7)-sqrt(31))/2
Find the zeros of x^2-sqrt(7)x-6=0
The zeros of x^2-sqrt(7)x-6=0 are x=(sqrt(7)+sqrt(31))/2 ,x=(sqrt(7)-sqrt(31))/2