Solve 30*(x-5)(x+4)geq0

$$30 \cdot ( x - 5 ) ( x + 4 ) \geq 0$$

Solve for x

$$x\in (-\infty,-4]\cup [5,\infty)$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 30 by x-5.

$$\left(30x-150\right)\left(x+4\right)\geq 0$$

Use the distributive property to multiply 30x-150 by x+4 and combine like terms.

$$30x^{2}-30x-600\geq 0$$

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.

$$30x^{2}-30x-600=0$$

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 30 for a, -30 for b, and -600 for c in the quadratic formula.

$$x=\frac{-\left(-30\right)±\sqrt{\left(-30\right)^{2}-4\times 30\left(-600\right)}}{2\times 30}$$

Do the calculations.

$$x=\frac{30±270}{60}$$

Solve the equation x=\frac{30±270}{60} when ± is plus and when ± is minus.

$$x=5 x=-4$$

Rewrite the inequality by using the obtained solutions.

$$30\left(x-5\right)\left(x+4\right)\geq 0$$

For the product to be ≥0, x-5 and x+4 have to be both ≤0 or both ≥0. Consider the case when x-5 and x+4 are both ≤0.

$$x-5\leq 0 x+4\leq 0$$

The solution satisfying both inequalities is x\leq -4.

$$x\leq -4$$

Consider the case when x-5 and x+4 are both ≥0.

$$x+4\geq 0 x-5\geq 0$$

The solution satisfying both inequalities is x\geq 5.

$$x\geq 5$$

The final solution is the union of the obtained solutions.

$$x\leq -4\text{; }x\geq 5$$