Solve x^2+14x-28leq0

$$x ^ { 2 } + 14 x - 28 \leq 0$$

Solve for x

$$x\in \begin{bmatrix}-\sqrt{77}-7,\sqrt{77}-7\end{bmatrix}$$

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.

$$x^{2}+14x-28=0$$

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, 14 for b, and -28 for c in the quadratic formula.

$$x=\frac{-14±\sqrt{14^{2}-4\times 1\left(-28\right)}}{2}$$

Do the calculations.

$$x=\frac{-14±2\sqrt{77}}{2}$$

Solve the equation x=\frac{-14±2\sqrt{77}}{2} when ± is plus and when ± is minus.

$$x=\sqrt{77}-7 x=-\sqrt{77}-7$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-\left(\sqrt{77}-7\right)\right)\left(x-\left(-\sqrt{77}-7\right)\right)\leq 0$$

For the product to be ≤0, one of the values x-\left(\sqrt{77}-7\right) and x-\left(-\sqrt{77}-7\right) has to be ≥0 and the other has to be ≤0. Consider the case when x-\left(\sqrt{77}-7\right)\geq 0 and x-\left(-\sqrt{77}-7\right)\leq 0.

$$x-\left(\sqrt{77}-7\right)\geq 0 x-\left(-\sqrt{77}-7\right)\leq 0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-\left(\sqrt{77}-7\right)\leq 0 and x-\left(-\sqrt{77}-7\right)\geq 0.

$$x-\left(-\sqrt{77}-7\right)\geq 0 x-\left(\sqrt{77}-7\right)\leq 0$$

The solution satisfying both inequalities is x\in \left[-\left(\sqrt{77}+7\right),\sqrt{77}-7\right].

$$x\in \begin{bmatrix}-\left(\sqrt{77}+7\right),\sqrt{77}-7\end{bmatrix}$$

The final solution is the union of the obtained solutions.

$$x\in \begin{bmatrix}-\sqrt{77}-7,\sqrt{77}-7\end{bmatrix}$$