What are the solutions to the equation 2x^2-5x=1 ?

The solutions to the equation 2x^2-5x=1 are x=(5+sqrt(33))/4 ,x=(5-sqrt(33))/4

Find the zeros of 2x^2-5x=1

The zeros of 2x^2-5x=1 are x=(5+sqrt(33))/4 ,x=(5-sqrt(33))/4

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2 x^{2} - 5 x = 1

x = \frac{5 + 33}{4}, x = \frac{5 - 33}{4}

Decimal Notation

x = 2.68614 \dots, x = - 0.18614 \dots

Solution steps

2 x^{2} - 5 x = 1

Move

1

to the left side

2 x^{2} - 5 x - 1 = 0

Solve with the quadratic formula

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

(- 5)^{2} - 4 \cdot 2 (- 1) = 33

x_{1, 2} = \frac{- ( - 5 ) \pm 33}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 33}{2 \cdot 2}, x_{2} = \frac{- ( - 5 ) - 33}{2 \cdot 2}

x = \frac{- ( - 5 ) + 33}{2 \cdot 2} : \frac{5 + 33}{4}

x = \frac{- ( - 5 ) - 33}{2 \cdot 2} : \frac{5 - 33}{4}

The solutions to the quadratic equation are:

x = \frac{5 + 33}{4}, x = \frac{5 - 33}{4}

2 u - 10 = 8

\frac{3 + \frac{a}{b}}{6 - \frac{a}{b}}

\frac{2}{16} = 8

2 x = 4 - x

\frac{7}{8} y - 6 = 8

What are the solutions to the equation 2x^2-5x=1 ?
The solutions to the equation 2x^2-5x=1 are x=(5+sqrt(33))/4 ,x=(5-sqrt(33))/4
Find the zeros of 2x^2-5x=1
The zeros of 2x^2-5x=1 are x=(5+sqrt(33))/4 ,x=(5-sqrt(33))/4