What are the solutions to the equation x^2-14x-4=0 ?

The solutions to the equation x^2-14x-4=0 are x=7+sqrt(53),x=7-sqrt(53)

Find the zeros of x^2-14x-4=0

The zeros of x^2-14x-4=0 are x=7+sqrt(53),x=7-sqrt(53)

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x^{2} - 14 x - 4 = 0

x = 7 + 53, x = 7 - 53

Decimal Notation

x = 14.28010 \dots, x = - 0.28010 \dots

Solution steps

x^{2} - 14 x - 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 14 ) \pm ( - 14 ) ^{2} - 4 \cdot 1 \cdot ( - 4 )}{2 \cdot 1}

(- 14)^{2} - 4 \cdot 1 \cdot (- 4) = 253

x_{1, 2} = \frac{- ( - 14 ) \pm 2 53}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 14 ) + 2 53}{2 \cdot 1}, x_{2} = \frac{- ( - 14 ) - 2 53}{2 \cdot 1}

x = \frac{- ( - 14 ) + 2 53}{2 \cdot 1} : 7 + 53

x = \frac{- ( - 14 ) - 2 53}{2 \cdot 1} : 7 - 53

The solutions to the quadratic equation are:

x = 7 + 53, x = 7 - 53

3 x + 4 \geq 22

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

0 = 16 t^{2} - 140 t + 68

\frac{1}{2} = \frac{4}{5} x - \frac{1}{4}

\frac{2}{x} - 3 \leq 5

What are the solutions to the equation x^2-14x-4=0 ?
The solutions to the equation x^2-14x-4=0 are x=7+sqrt(53),x=7-sqrt(53)
Find the zeros of x^2-14x-4=0
The zeros of x^2-14x-4=0 are x=7+sqrt(53),x=7-sqrt(53)