Solve x^2+432+43x>0

$$x ^ { 2 } + 432 + 43 x > 0$$

Solve for x

$$x\in \left(-\infty,-27\right)\cup \left(-16,\infty\right)$$

Steps Using the Quadratic Formula

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.

$$x^{2}+432+43x=0$$

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, 43 for b, and 432 for c in the quadratic formula.

$$x=\frac{-43±\sqrt{43^{2}-4\times 1\times 432}}{2}$$

Do the calculations.

$$x=\frac{-43±11}{2}$$

Solve the equation x=\frac{-43±11}{2} when ± is plus and when ± is minus.

$$x=-16 x=-27$$

Rewrite the inequality by using the obtained solutions.

$$\left(x+16\right)\left(x+27\right)>0$$

For the product to be positive, x+16 and x+27 have to be both negative or both positive. Consider the case when x+16 and x+27 are both negative.

$$x+16<0 x+27<0$$

The solution satisfying both inequalities is x<-27.

$$x<-27$$

Consider the case when x+16 and x+27 are both positive.

$$x+27>0 x+16>0$$

The solution satisfying both inequalities is x>-16.

$$x>-16$$

The final solution is the union of the obtained solutions.

$$x<-27\text{; }x>-16$$