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

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

Solve for x

$$x\in \left(-\infty,-4\right)\cup \left(1,\infty\right)$$

Steps Using the Quadratic Formula

Multiply the inequality by -1 to make the coefficient of the highest power in -x^{2}-3x+4 positive. Since -1 is negative, the inequality direction is changed.

$$x^{2}+3x-4>0$$

To solve the inequality, factor the left hand side. Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$x^{2}+3x-4=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, 3 for b, and -4 for c in the quadratic formula.

$$x=\frac{-3±\sqrt{3^{2}-4\times 1\left(-4\right)}}{2}$$

Do the calculations.

$$x=\frac{-3±5}{2}$$

Solve the equation x=\frac{-3±5}{2} when ± is plus and when ± is minus.

$$x=1 x=-4$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-1\right)\left(x+4\right)>0$$

For the product to be positive, x-1 and x+4 have to be both negative or both positive. Consider the case when x-1 and x+4 are both negative.

$$x-1<0 x+4<0$$

The solution satisfying both inequalities is x<-4.

$$x<-4$$

Consider the case when x-1 and x+4 are both positive.

$$x+4>0 x-1>0$$

The solution satisfying both inequalities is x>1.

$$x>1$$

The final solution is the union of the obtained solutions.

$$x<-4\text{; }x>1$$