Solve x(x-3)-4(5-x)geq0

$$x ( x - 3 ) - 4 ( 5 - x ) \geq 0$$

Solve for x

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

Steps Using the Quadratic Formula

Use the distributive property to multiply x by x-3.

$$x^{2}-3x-4\left(5-x\right)\geq 0$$

Use the distributive property to multiply -4 by 5-x.

$$x^{2}-3x-20+4x\geq 0$$

Combine -3x and 4x to get x.

$$x^{2}+x-20\geq 0$$

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.

$$x^{2}+x-20=0$$

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, 1 for b, and -20 for c in the quadratic formula.

$$x=\frac{-1±\sqrt{1^{2}-4\times 1\left(-20\right)}}{2}$$

Do the calculations.

$$x=\frac{-1±9}{2}$$

Solve the equation x=\frac{-1±9}{2} when ± is plus and when ± is minus.

$$x=4 x=-5$$

Rewrite the inequality by using the obtained solutions.

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

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

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

The solution satisfying both inequalities is x\leq -5.

$$x\leq -5$$

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

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

The solution satisfying both inequalities is x\geq 4.

$$x\geq 4$$

The final solution is the union of the obtained solutions.

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