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

The solutions to the equation x^2+(3/2 x-1/4)=0 are x=(-3+sqrt(13))/4 ,x=(-3-sqrt(13))/4

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

The zeros of x^2+(3/2 x-1/4)=0 are x=(-3+sqrt(13))/4 ,x=(-3-sqrt(13))/4

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

x = \frac{- 3 + 13}{4}, x = \frac{- 3 - 13}{4}

Decimal Notation

x = 0.15138 \dots, x = - 1.65138 \dots

Solution steps

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

Refine

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

Write in the standard form

a x^{2} + b x + c = 0

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

Solve with the quadratic formula

x_{1, 2} = \frac{- \frac{3}{2} \pm ( \frac{3}{2} ) ^{2} - 4 \cdot 1 \cdot ( - \frac{1}{4} )}{2 \cdot 1}

(\frac{3}{2})^{2} - 4 \cdot 1 \cdot (- \frac{1}{4}) = \frac{13}{2}

x_{1, 2} = \frac{- \frac{3}{2} \pm \frac{13}{2}}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- \frac{3}{2} + \frac{13}{2}}{2 \cdot 1}, x_{2} = \frac{- \frac{3}{2} - \frac{13}{2}}{2 \cdot 1}

x = \frac{- \frac{3}{2} + \frac{13}{2}}{2 \cdot 1} : \frac{- 3 + 13}{4}

x = \frac{- \frac{3}{2} - \frac{13}{2}}{2 \cdot 1} : \frac{- 3 - 13}{4}

The solutions to the quadratic equation are:

x = \frac{- 3 + 13}{4}, x = \frac{- 3 - 13}{4}

3 (\frac{2}{3})^{3}

\frac{2 2}{3 - 2}

\frac{( 3 ) ^{3}}{( 3 ) ^{4}}

\frac{7}{9} - \frac{2}{3}

3 - \frac{x}{2} \leq 18

What are the solutions to the equation x^2+(3/2 x-1/4)=0 ?
The solutions to the equation x^2+(3/2 x-1/4)=0 are x=(-3+sqrt(13))/4 ,x=(-3-sqrt(13))/4
Find the zeros of x^2+(3/2 x-1/4)=0
The zeros of x^2+(3/2 x-1/4)=0 are x=(-3+sqrt(13))/4 ,x=(-3-sqrt(13))/4