Solve 6x^2+5x-3geq0

$$6 x ^ { 2 } + 5 x - 3 \geq 0$$

Solve for x

$$x\in (-\infty,\frac{-\sqrt{97}-5}{12}]\cup [\frac{\sqrt{97}-5}{12},\infty)$$

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.

$$6x^{2}+5x-3=0$$

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 6 for a, 5 for b, and -3 for c in the quadratic formula.

$$x=\frac{-5±\sqrt{5^{2}-4\times 6\left(-3\right)}}{2\times 6}$$

Do the calculations.

$$x=\frac{-5±\sqrt{97}}{12}$$

Solve the equation x=\frac{-5±\sqrt{97}}{12} when ± is plus and when ± is minus.

$$x=\frac{\sqrt{97}-5}{12} x=\frac{-\sqrt{97}-5}{12}$$

Rewrite the inequality by using the obtained solutions.

$$6\left(x-\frac{\sqrt{97}-5}{12}\right)\left(x-\frac{-\sqrt{97}-5}{12}\right)\geq 0$$

For the product to be ≥0, x-\frac{\sqrt{97}-5}{12} and x-\frac{-\sqrt{97}-5}{12} have to be both ≤0 or both ≥0. Consider the case when x-\frac{\sqrt{97}-5}{12} and x-\frac{-\sqrt{97}-5}{12} are both ≤0.

$$x-\frac{\sqrt{97}-5}{12}\leq 0 x-\frac{-\sqrt{97}-5}{12}\leq 0$$

The solution satisfying both inequalities is x\leq \frac{-\sqrt{97}-5}{12}.

$$x\leq \frac{-\sqrt{97}-5}{12}$$

Consider the case when x-\frac{\sqrt{97}-5}{12} and x-\frac{-\sqrt{97}-5}{12} are both ≥0.

$$x-\frac{-\sqrt{97}-5}{12}\geq 0 x-\frac{\sqrt{97}-5}{12}\geq 0$$

The solution satisfying both inequalities is x\geq \frac{\sqrt{97}-5}{12}.

$$x\geq \frac{\sqrt{97}-5}{12}$$

The final solution is the union of the obtained solutions.

$$x\leq \frac{-\sqrt{97}-5}{12}\text{; }x\geq \frac{\sqrt{97}-5}{12}$$