Solve 15x^2+4x-3geq0

$$15 x ^ { 2 } + 4 x - 3 \geq 0$$

Solve for x

$$x\in (-\infty,-\frac{3}{5}]\cup [\frac{1}{3},\infty)$$

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.

$$15x^{2}+4x-3=0$$

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 15 for a, 4 for b, and -3 for c in the quadratic formula.

$$x=\frac{-4±\sqrt{4^{2}-4\times 15\left(-3\right)}}{2\times 15}$$

Do the calculations.

$$x=\frac{-4±14}{30}$$

Solve the equation x=\frac{-4±14}{30} when ± is plus and when ± is minus.

$$x=\frac{1}{3} x=-\frac{3}{5}$$

Rewrite the inequality by using the obtained solutions.

$$15\left(x-\frac{1}{3}\right)\left(x+\frac{3}{5}\right)\geq 0$$

For the product to be ≥0, x-\frac{1}{3} and x+\frac{3}{5} have to be both ≤0 or both ≥0. Consider the case when x-\frac{1}{3} and x+\frac{3}{5} are both ≤0.

$$x-\frac{1}{3}\leq 0 x+\frac{3}{5}\leq 0$$

The solution satisfying both inequalities is x\leq -\frac{3}{5}.

$$x\leq -\frac{3}{5}$$

Consider the case when x-\frac{1}{3} and x+\frac{3}{5} are both ≥0.

$$x+\frac{3}{5}\geq 0 x-\frac{1}{3}\geq 0$$

The solution satisfying both inequalities is x\geq \frac{1}{3}.

$$x\geq \frac{1}{3}$$

The final solution is the union of the obtained solutions.

$$x\leq -\frac{3}{5}\text{; }x\geq \frac{1}{3}$$