What is 50<= 0.004x^2-x+80 ?

The solution to 50 = 25sqrt(13)+125

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50<= 0.004x^2-x+80

50 \leq 0.004 x^{2} - x + 80

x \leq - 2513 + 125 or x \geq 2513 + 125

Interval Notation

(- \infty, - 2513 + 125] \cup [2513 + 125, \infty)

Decimal Notation

x \leq 34.86121 \dots or x \geq 215.13878 \dots

Solution steps

50 \leq 0.004 x^{2} - x + 80

Switch sides

0.004 x^{2} - x + 80 \geq 50

Rewrite in standard form

0.004 x^{2} - x + 30 \geq 0

Complete the square

0.004 x^{2} - x + 30 : 0.004 (x - 125)^{2} - 32.5

0.004 (x - 125)^{2} - 32.5 \geq 0

Multiply both sides by

1000

4 (x - 125)^{2} - 32500 \geq 0

Move

32500

to the right side

4 (x - 125)^{2} \geq 32500

Divide both sides by

4

(x - 125)^{2} \geq 8125

For

u^{n} \geq a

, if

n

is even then

x - 125 \leq - 8125 or x - 125 \geq 8125

x - 125 \leq - 8125 : x \leq - 2513 + 125

x - 125 \geq 8125 : x \geq 2513 + 125

Combine the intervals

x \leq - 2513 + 125 or x \geq 2513 + 125

((\frac{- 2}{6})^{- 4})^{2}

- 2 x^{2} + 122 = 24

63

\frac{∣ x + 6 ∣}{8} \leq 1

6 y^{2} - 24 y = 0

What is 50<= 0.004x^2-x+80 ?
The solution to 50<= 0.004x^2-x+80 is x<=-25sqrt(13)+125\lor x>= 25sqrt(13)+125