Solve 5x^2+19x-4leq0

$$5 x ^ { 2 } + 19 x - 4 \leq 0$$

Solve for x

$$x\in \begin{bmatrix}-4,\frac{1}{5}\end{bmatrix}$$

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.

$$5x^{2}+19x-4=0$$

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 5 for a, 19 for b, and -4 for c in the quadratic formula.

$$x=\frac{-19±\sqrt{19^{2}-4\times 5\left(-4\right)}}{2\times 5}$$

Do the calculations.

$$x=\frac{-19±21}{10}$$

Solve the equation x=\frac{-19±21}{10} when ± is plus and when ± is minus.

$$x=\frac{1}{5} x=-4$$

Rewrite the inequality by using the obtained solutions.

$$5\left(x-\frac{1}{5}\right)\left(x+4\right)\leq 0$$

For the product to be ≤0, one of the values x-\frac{1}{5} and x+4 has to be ≥0 and the other has to be ≤0. Consider the case when x-\frac{1}{5}\geq 0 and x+4\leq 0.

$$x-\frac{1}{5}\geq 0 x+4\leq 0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-\frac{1}{5}\leq 0 and x+4\geq 0.

$$x+4\geq 0 x-\frac{1}{5}\leq 0$$

The solution satisfying both inequalities is x\in \left[-4,\frac{1}{5}\right].

$$x\in \begin{bmatrix}-4,\frac{1}{5}\end{bmatrix}$$

The final solution is the union of the obtained solutions.

$$x\in \begin{bmatrix}-4,\frac{1}{5}\end{bmatrix}$$