How do solve $x^2<10-3x$ and write the answer as a inequality and interval notation?

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Answer 1 · 2017-01-31T14:14:42

The answer is $-5 < x <2$
$x in ]-5, 2[$

Let's rewrite the inequality

$x^2+3x-10<0$

Let's factorise

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

Let $f(x)=(x-2)(x+5)$

We can build the sign chart

$color(white)(aaaa)$ $x$ $color(white)(aaaa)$ $-oo$ $color(white)(aaaa)$ $-5$ $color(white)(aaaa)$ $2$ $color(white)(aaaaa)$ $+oo$

$color(white)(aaaa)$ $x+5$ $color(white)(aaaaa)$ $-$ $color(white)(aaaa)$ $+$ $color(white)(aaaa)$ $+$

$color(white)(aaaa)$ $x-2$ $color(white)(aaaaa)$ $-$ $color(white)(aaaa)$ $-$ $color(white)(aaaa)$ $+$

$color(white)(aaaa)$ $f(x)$ $color(white)(aaaaaa)$ $+$ $color(white)(aaaa)$ $-$ $color(white)(aaaa)$ $+$

Therefore,

$f(x)<0$ when $x in ]-5, 2[$

or $-5 < x <2$

How do solve x^2<10-3xx^2<10-3x and write the answer as a inequality and interval notation?