Solve the Quadratic Inequality: -x² + 3x + 4 < 0

Solve the following equation:

$-x^2+3x+4<0$

To solve the inequality $-x^2 + 3x + 4 < 0$, we begin by finding the roots of the equation $-x^2 + 3x + 4 = 0$.

Step 1: Calculate the discriminant using $b^2 - 4ac$ with $a = -1$, $b = 3$, and $c = 4$:
$\Delta = 3^2 - 4 \times (-1) \times 4 = 9 + 16 = 25$

Step 2: Compute the roots using the quadratic formula:
$x = \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{-3 \pm \sqrt{25}}{-2}$

Calculating the roots:
First root: $x_1 = \frac{-3 + 5}{-2} = -1$
Second root: $x_2 = \frac{-3 - 5}{-2} = 4$

Step 3: Analyze intervals defined by the roots $-1$ and $4$. Given that the parabola opens downwards, we check intervals $(-\infty, -1)$, $(-1, 4)$, and $(4, \infty)$.

Testing a point in each interval to determine the sign:
Interval $(-\infty, -1)$, test $x = -2$: $-(-2)^2 + 3(-2) + 4 = -4 - 6 + 4 = -6$ (negative)
Interval $(-1, 4)$, test $x = 0$: $-0^2 + 3(0) + 4 = 4$ (positive)
Interval $(4, \infty)$, test $x = 5$: $-(5)^2 + 3(5) + 4 = -25 + 15 + 4 = -6$ (negative)

From the test results, the quadratic expression is negative in the intervals $(-\infty, -1)$ and $(4, \infty)$.

Therefore, the solution to the inequality $-x^2 + 3x + 4 < 0$ is $x < -1$ or $4 < x$.