What is 2x^2-3x<= 1-5x ?

The solution to 2x^2-3x<= 1-5x is (-sqrt(3)-1)/2 <= x<= (sqrt(3)-1)/2

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2x^2-3x<= 1-5x

2 x^{2} - 3 x \leq 1 - 5 x

\frac{- 3 - 1}{2} \leq x \leq \frac{3 - 1}{2}

Interval Notation

[\frac{- 3 - 1}{2}, \frac{3 - 1}{2}]

Decimal Notation

- 1.36602 \dots \leq x \leq 0.36602 \dots

Solution steps

2 x^{2} - 3 x \leq 1 - 5 x

Rewrite in standard form

2 x^{2} + 2 x - 1 \leq 0

Complete the square

2 x^{2} + 2 x - 1 : 2 (x + \frac{1}{2})^{2} - \frac{3}{2}

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

Move

\frac{3}{2}

to the right side

2 (x + \frac{1}{2})^{2} \leq \frac{3}{2}

Divide both sides by

2

(x + \frac{1}{2})^{2} \leq \frac{3}{4}

For

u^{n} \leq a

, if

n

is even then

- \frac{3}{4} \leq x + \frac{1}{2} \leq \frac{3}{4}

a \leq u \leq b

then

a \leq u and u \leq b

- \frac{3}{4} \leq x + \frac{1}{2} and x + \frac{1}{2} \leq \frac{3}{4}

- \frac{3}{4} \leq x + \frac{1}{2} : x \geq \frac{- 3 - 1}{2}

x + \frac{1}{2} \leq \frac{3}{4} : x \leq \frac{3 - 1}{2}

Combine the intervals

x \geq \frac{- 3 - 1}{2} and x \leq \frac{3 - 1}{2}

Merge Overlapping Intervals

\frac{- 3 - 1}{2} \leq x \leq \frac{3 - 1}{2}

∣ 4 x - 6 ∣ = 34

f = 1^{3} - 9 \cdot 1

z^{2} + 16 = 0

\frac{x}{x + 4} \geq 2

125 b^{2} - 20 c^{4}

What is 2x^2-3x<= 1-5x ?
The solution to 2x^2-3x<= 1-5x is (-sqrt(3)-1)/2 <= x<= (sqrt(3)-1)/2