How do you solve $2 x^{2} + x < 15$ $2x^2+x<15$ ?

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Answer 1 · 2018-06-25T16:03:44

$- 3 < x < 2.5$ $-3 < x <2.5$

We can re-arrange and say:

$2 x^{2} + x - 15 < 0$ $2x^2+x-15<0$

By factorising:

$(2 x - 5) (x + 3) < 0$ $(2x-5)(x+3)<0$

$therefore$ $2x-5<0$

$rArr 2x<5$

$rArr x<2.5$

$x+3<0$

$rArr x<-3$

Therefore:

$therefore$ $-3 < x <2.5$

How do you solve 2x^2+x<152x2+x<152x^2+x<15?