What is -3x^2+16x>x+12 ?

The solution to -3x^2+16x>x+12 is 1<x<4

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-3x^2+16x>x+12

- 3 x^{2} + 16 x > x + 12

1 < x < 4

Interval Notation

(1, 4)

Solution steps

- 3 x^{2} + 16 x > x + 12

Rewrite in standard form

- x^{2} + 5 x - 4 > 0

Factor

- x^{2} + 5 x - 4 : - (x - 1) (x - 4)

- (x - 1) (x - 4) > 0

Multiply both sides by

- 1

(reverse the inequality)

(- (x - 1) (x - 4)) (- 1) < 0 \cdot (- 1)

Simplify

(x - 1) (x - 4) < 0

Identify the intervals

1 < x < 4

- 4 x^{2} + 8 x = 11

(- 243)^{\frac{1}{5}}

3 w^{2} - 4 w + 5 = 0

625 x - 100 x^{2} + 4 x^{3} > 1000

2 a^{2} b - 4 a b^{2}