What are the solutions to the equation x^2-x-0.25=0 ?

The solutions to the equation x^2-x-0.25=0 are x=(1+sqrt(2))/2 ,x=(1-sqrt(2))/2

Find the zeros of x^2-x-0.25=0

The zeros of x^2-x-0.25=0 are x=(1+sqrt(2))/2 ,x=(1-sqrt(2))/2

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x^{2} - x - 0.25 = 0

x = \frac{1 + 2}{2}, x = \frac{1 - 2}{2}

Decimal Notation

x = 1.20710 \dots, x = - 0.20710 \dots

Solution steps

x^{2} - x - 0.25 = 0

Multiply both sides by

100

100 x^{2} - 100 x - 25 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 100 ) \pm ( - 100 ) ^{2} - 4 \cdot 100 ( - 25 )}{2 \cdot 100}

(- 100)^{2} - 4 \cdot 100 (- 25) = 1002

x_{1, 2} = \frac{- ( - 100 ) \pm 100 2}{2 \cdot 100}

Separate the solutions

x_{1} = \frac{- ( - 100 ) + 100 2}{2 \cdot 100}, x_{2} = \frac{- ( - 100 ) - 100 2}{2 \cdot 100}

x = \frac{- ( - 100 ) + 100 2}{2 \cdot 100} : \frac{1 + 2}{2}

x = \frac{- ( - 100 ) - 100 2}{2 \cdot 100} : \frac{1 - 2}{2}

The solutions to the quadratic equation are:

x = \frac{1 + 2}{2}, x = \frac{1 - 2}{2}

\frac{4 + 10 i}{3 + i}

2 = c^{2} + c

\frac{n}{12} = - 5 - 11

\frac{m + 4}{3} > 3

144 x^{3} y^{4} z^{5}

What are the solutions to the equation x^2-x-0.25=0 ?
The solutions to the equation x^2-x-0.25=0 are x=(1+sqrt(2))/2 ,x=(1-sqrt(2))/2
Find the zeros of x^2-x-0.25=0
The zeros of x^2-x-0.25=0 are x=(1+sqrt(2))/2 ,x=(1-sqrt(2))/2