What is 2x^2-6x+3<= 0 ?

The solution to 2x^2-6x+3<= 0 is (-sqrt(3)+3)/2 <= x<= (sqrt(3)+3)/2

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2x^2-6x+3<= 0

2 x^{2} - 6 x + 3 \leq 0

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

Interval Notation

[\frac{- 3 + 3}{2}, \frac{3 + 3}{2}]

Decimal Notation

0.63397 \dots \leq x \leq 2.36602 \dots

Solution steps

2 x^{2} - 6 x + 3 \leq 0

Complete the square

2 x^{2} - 6 x + 3 : 2 (x - \frac{3}{2})^{2} - \frac{3}{2}

2 (x - \frac{3}{2})^{2} - \frac{3}{2} \leq 0

Move

\frac{3}{2}

to the right side

2 (x - \frac{3}{2})^{2} \leq \frac{3}{2}

Divide both sides by

2

(x - \frac{3}{2})^{2} \leq \frac{3}{4}

For

u^{n} \leq a

, if

n

is even then

- \frac{3}{4} \leq x - \frac{3}{2} \leq \frac{3}{4}

a \leq u \leq b

then

a \leq u and u \leq b

- \frac{3}{4} \leq x - \frac{3}{2} and x - \frac{3}{2} \leq \frac{3}{4}

- \frac{3}{4} \leq x - \frac{3}{2} : x \geq \frac{- 3 + 3}{2}

x - \frac{3}{2} \leq \frac{3}{4} : x \leq \frac{3 + 3}{2}

Combine the intervals

x \geq \frac{- 3 + 3}{2} and x \leq \frac{3 + 3}{2}

Merge Overlapping Intervals

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

1 - \frac{12}{52}

\frac{6}{7} x - \frac{1}{8} \leq \frac{11}{14} x - \frac{8}{7}

3 x - 4 = - 13

\frac{5 x}{4} + \frac{x}{3} = - 2

π x^{2} - x - 29 π = 0

What is 2x^2-6x+3<= 0 ?
The solution to 2x^2-6x+3<= 0 is (-sqrt(3)+3)/2 <= x<= (sqrt(3)+3)/2