What is 2x^2+5x>=-x^2+1 ?

The solution to 2x^2+5x>=-x^2+1 is x = (sqrt(37)-5)/6

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2x^2+5x>=-x^2+1

2 x^{2} + 5 x \geq - x^{2} + 1

x \leq \frac{- 37 - 5}{6} or x \geq \frac{37 - 5}{6}

Interval Notation

(- \infty, \frac{- 37 - 5}{6}] \cup [\frac{37 - 5}{6}, \infty)

Decimal Notation

x \leq - 1.84712 \dots or x \geq 0.18046 \dots

Solution steps

2 x^{2} + 5 x \geq - x^{2} + 1

Rewrite in standard form

3 x^{2} + 5 x - 1 \geq 0

Complete the square

3 x^{2} + 5 x - 1 : 3 (x + \frac{5}{6})^{2} - \frac{37}{12}

3 (x + \frac{5}{6})^{2} - \frac{37}{12} \geq 0

Move

\frac{37}{12}

to the right side

3 (x + \frac{5}{6})^{2} \geq \frac{37}{12}

Divide both sides by

3

(x + \frac{5}{6})^{2} \geq \frac{37}{36}

For

u^{n} \geq a

, if

n

is even then

x + \frac{5}{6} \leq - \frac{37}{36} or x + \frac{5}{6} \geq \frac{37}{36}

x + \frac{5}{6} \leq - \frac{37}{36} : x \leq \frac{- 37 - 5}{6}

x + \frac{5}{6} \geq \frac{37}{36} : x \geq \frac{37 - 5}{6}

Combine the intervals

x \leq \frac{- 37 - 5}{6} or x \geq \frac{37 - 5}{6}

x - 4 = 16

- \frac{3}{7} - \frac{1}{2}

0 = 85.5 - 0.1702 x

4 - 4

\frac{5 x + 10}{2} = 5

What is 2x^2+5x>=-x^2+1 ?
The solution to 2x^2+5x>=-x^2+1 is x<= (-sqrt(37)-5)/6 \lor x>= (sqrt(37)-5)/6