What are the solutions to the equation 2x^2-7x-1=0 ?

The solutions to the equation 2x^2-7x-1=0 are x=(7+sqrt(57))/4 ,x=(7-sqrt(57))/4

Find the zeros of 2x^2-7x-1=0

The zeros of 2x^2-7x-1=0 are x=(7+sqrt(57))/4 ,x=(7-sqrt(57))/4

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2 x^{2} - 7 x - 1 = 0

x = \frac{7 + 57}{4}, x = \frac{7 - 57}{4}

Decimal Notation

x = 3.63745 \dots, x = - 0.13745 \dots

Solution steps

2 x^{2} - 7 x - 1 = 0

Solve with the quadratic formula

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

(- 7)^{2} - 4 \cdot 2 (- 1) = 57

x_{1, 2} = \frac{- ( - 7 ) \pm 57}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 7 ) + 57}{2 \cdot 2}, x_{2} = \frac{- ( - 7 ) - 57}{2 \cdot 2}

x = \frac{- ( - 7 ) + 57}{2 \cdot 2} : \frac{7 + 57}{4}

x = \frac{- ( - 7 ) - 57}{2 \cdot 2} : \frac{7 - 57}{4}

The solutions to the quadratic equation are:

x = \frac{7 + 57}{4}, x = \frac{7 - 57}{4}

4 k - 7 (5 k - 2) \geq 169

1 - 8 j > 9

\frac{π}{3} \cdot \frac{180}{π}

5 x^{2} + 21 x - 20

\frac{\frac{16}{5}}{( - \frac{12}{25} )}

What are the solutions to the equation 2x^2-7x-1=0 ?
The solutions to the equation 2x^2-7x-1=0 are x=(7+sqrt(57))/4 ,x=(7-sqrt(57))/4
Find the zeros of 2x^2-7x-1=0
The zeros of 2x^2-7x-1=0 are x=(7+sqrt(57))/4 ,x=(7-sqrt(57))/4