Solve 6x+55-x^2geq0

$$6 x + 55 - x ^ { 2 } \geq 0$$

Solve for x

$$x\in \begin{bmatrix}-5,11\end{bmatrix}$$

Steps Using the Quadratic Formula

Multiply the inequality by -1 to make the coefficient of the highest power in 6x+55-x^{2} positive. Since -1 is negative, the inequality direction is changed.

$$-6x-55+x^{2}\leq 0$$

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.

$$-6x-55+x^{2}=0$$

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, -6 for b, and -55 for c in the quadratic formula.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 1\left(-55\right)}}{2}$$

Do the calculations.

$$x=\frac{6±16}{2}$$

Solve the equation x=\frac{6±16}{2} when ± is plus and when ± is minus.

$$x=11 x=-5$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-11\right)\left(x+5\right)\leq 0$$

For the product to be ≤0, one of the values x-11 and x+5 has to be ≥0 and the other has to be ≤0. Consider the case when x-11\geq 0 and x+5\leq 0.

$$x-11\geq 0 x+5\leq 0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-11\leq 0 and x+5\geq 0.

$$x+5\geq 0 x-11\leq 0$$

The solution satisfying both inequalities is x\in \left[-5,11\right].

$$x\in \begin{bmatrix}-5,11\end{bmatrix}$$

The final solution is the union of the obtained solutions.

$$x\in \begin{bmatrix}-5,11\end{bmatrix}$$