What are the solutions to the equation x^2-5x-7=0 ?

The solutions to the equation x^2-5x-7=0 are x=(5+sqrt(53))/2 ,x=(5-sqrt(53))/2

Find the zeros of x^2-5x-7=0

The zeros of x^2-5x-7=0 are x=(5+sqrt(53))/2 ,x=(5-sqrt(53))/2

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x^{2} - 5 x - 7 = 0

x = \frac{5 + 53}{2}, x = \frac{5 - 53}{2}

Decimal Notation

x = 6.14005 \dots, x = - 1.14005 \dots

Solution steps

x^{2} - 5 x - 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 1 \cdot ( - 7 )}{2 \cdot 1}

(- 5)^{2} - 4 \cdot 1 \cdot (- 7) = 53

x_{1, 2} = \frac{- ( - 5 ) \pm 53}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 53}{2 \cdot 1}, x_{2} = \frac{- ( - 5 ) - 53}{2 \cdot 1}

x = \frac{- ( - 5 ) + 53}{2 \cdot 1} : \frac{5 + 53}{2}

x = \frac{- ( - 5 ) - 53}{2 \cdot 1} : \frac{5 - 53}{2}

The solutions to the quadratic equation are:

x = \frac{5 + 53}{2}, x = \frac{5 - 53}{2}

x^{3} - 4 x^{2} - x + 4

0 = x + 23 \cdot 5 - 4.9 \cdot 5^{2}

11 c^{2} - 7 c^{3}

(2 y^{2}) (4 x y^{3}) + (3 x y^{4}) (5 y)

z^{2} - 3.25 z

What are the solutions to the equation x^2-5x-7=0 ?
The solutions to the equation x^2-5x-7=0 are x=(5+sqrt(53))/2 ,x=(5-sqrt(53))/2
Find the zeros of x^2-5x-7=0
The zeros of x^2-5x-7=0 are x=(5+sqrt(53))/2 ,x=(5-sqrt(53))/2