# How do you solve -2x^2+6X+15 <=0?

Jun 5, 2016

Half closed intervals: (-inf., x2] and [x1, inf.)

#### Explanation:

$f \left(x\right) - 2 {x}^{2} + 6 x + 15 \le 0$
First, solve f(x) = 0 to find the 2 real roots. Use the improved quadratic formula (Socratic Search)
$D = {d}^{2} = {b}^{2} - 4 a c = 36 + 120 = 156$ --> $d = 2 \sqrt{39}$.
There are 2 real roots (2 x-intercepts):
$x = - \frac{b}{2 a} \pm \frac{d}{2 a} = \frac{6}{-} 4 \pm \frac{\sqrt{39}}{2} = \frac{3 \pm \sqrt{39}}{2}$
$x 1 = \frac{3 + \sqrt{39}}{2}$ and $x 2 = \frac{3 - \sqrt{39}}{2}$.
Since a < 0, the parabola opens downward.
f(x) > 0 inside the interval (x2, x1), between the 2 x-intercepts.
f(x) < 0 outside the (x2, x1) interval, because the 2 parabola branches are below the x-axis.
Solution set: Two half-closed intervals (-inf., x2] and [x1, inf.)
The 2 end points x1 and x2 are include in the solution set
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