What are the solutions to the equation x^2-3x+1/4 =0 ?

The solutions to the equation x^2-3x+1/4 =0 are x=(3+2sqrt(2))/2 ,x=(3-2sqrt(2))/2

Find the zeros of x^2-3x+1/4 =0

The zeros of x^2-3x+1/4 =0 are x=(3+2sqrt(2))/2 ,x=(3-2sqrt(2))/2

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x^{2} - 3 x + \frac{1}{4} = 0

x = \frac{3 + 2 2}{2}, x = \frac{3 - 2 2}{2}

Decimal

x = 2.91421 \dots, x = 0.08578 \dots

Solution steps

x^{2} - 3 x + \frac{1}{4} = 0

Multiply both sides by

4

4 x^{2} - 12 x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 4 \cdot 1}{2 \cdot 4}

(- 12)^{2} - 4 \cdot 4 \cdot 1 = 82

x_{1, 2} = \frac{- ( - 12 ) \pm 8 2}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 12 ) + 8 2}{2 \cdot 4}, x_{2} = \frac{- ( - 12 ) - 8 2}{2 \cdot 4}

x = \frac{- ( - 12 ) + 8 2}{2 \cdot 4} : \frac{3 + 2 2}{2}

x = \frac{- ( - 12 ) - 8 2}{2 \cdot 4} : \frac{3 - 2 2}{2}

The solutions to the quadratic equation are:

x = \frac{3 + 2 2}{2}, x = \frac{3 - 2 2}{2}

y, 2 x + 2 x y = 2 y + x^{2}

x, x^{2} y + 3 x y^{3} - x = 3

9 x^{2} - 4 x = 1

9 u^{2} - u - 6534 = 0

y, x^{2} = \frac{1}{y + 4}

What are the solutions to the equation x^2-3x+1/4 =0 ?
The solutions to the equation x^2-3x+1/4 =0 are x=(3+2sqrt(2))/2 ,x=(3-2sqrt(2))/2
Find the zeros of x^2-3x+1/4 =0
The zeros of x^2-3x+1/4 =0 are x=(3+2sqrt(2))/2 ,x=(3-2sqrt(2))/2