What is 3x^2+16x>6x+4 ?

The solution to 3x^2+16x>6x+4 is x (sqrt(37)-5)/3

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3x^2+16x>6x+4

3 x^{2} + 16 x > 6 x + 4

x < \frac{- 37 - 5}{3} or x > \frac{37 - 5}{3}

Interval Notation

(- \infty, \frac{- 37 - 5}{3}) \cup (\frac{37 - 5}{3}, \infty)

Decimal Notation

x < - 3.69425 \dots or x > 0.36092 \dots

Solution steps

3 x^{2} + 16 x > 6 x + 4

Rewrite in standard form

3 x^{2} + 10 x - 4 > 0

Complete the square

3 x^{2} + 10 x - 4 : 3 (x + \frac{5}{3})^{2} - \frac{37}{3}

3 (x + \frac{5}{3})^{2} - \frac{37}{3} > 0

Move

\frac{37}{3}

to the right side

3 (x + \frac{5}{3})^{2} > \frac{37}{3}

Divide both sides by

3

(x + \frac{5}{3})^{2} > \frac{37}{9}

For

u^{n} > a

, if

n

is even then

x + \frac{5}{3} < - \frac{37}{9} or x + \frac{5}{3} > \frac{37}{9}

x + \frac{5}{3} < - \frac{37}{9} : x < \frac{- 37 - 5}{3}

x + \frac{5}{3} > \frac{37}{9} : x > \frac{37 - 5}{3}

Combine the intervals

x < \frac{- 37 - 5}{3} or x > \frac{37 - 5}{3}

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

e^{- 20}

98 w^{2} = 4

\frac{7}{10 - 2}

2 x^{2} + 8 = 12 x

What is 3x^2+16x>6x+4 ?
The solution to 3x^2+16x>6x+4 is x<(-sqrt(37)-5)/3 \lor x>(sqrt(37)-5)/3