How do you solve -x^2-4x>3?

Sep 27, 2015

$- 3 < x < - 1$

Explanation:

$- {x}^{2} - 4 x > 3 \implies$ multiply by -1, reverse direction of inequality:
${x}^{2} + 4 x < - 3 \implies$ solving by completing the square:
${x}^{2} + 4 x + 4 < 4 - 3$
${\left(x + 2\right)}^{2} < 1$
$- 1 < \left(x + 2\right) < 1$
$- 3 < x < - 1$
In interval form:
$\left(- 3 , - 1\right)$