Solve 2x^2+5x-3geq0

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

Solve for x

$$x\in (-\infty,-3]\cup [\frac{1}{2},\infty)$$

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.

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

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

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

Do the calculations.

$$x=\frac{-5±7}{4}$$

Solve the equation x=\frac{-5±7}{4} when ± is plus and when ± is minus.

$$x=\frac{1}{2} x=-3$$

Rewrite the inequality by using the obtained solutions.

$$2\left(x-\frac{1}{2}\right)\left(x+3\right)\geq 0$$

For the product to be ≥0, x-\frac{1}{2} and x+3 have to be both ≤0 or both ≥0. Consider the case when x-\frac{1}{2} and x+3 are both ≤0.

$$x-\frac{1}{2}\leq 0 x+3\leq 0$$

The solution satisfying both inequalities is x\leq -3.

$$x\leq -3$$

Consider the case when x-\frac{1}{2} and x+3 are both ≥0.

$$x+3\geq 0 x-\frac{1}{2}\geq 0$$

The solution satisfying both inequalities is x\geq \frac{1}{2}.

$$x\geq \frac{1}{2}$$

The final solution is the union of the obtained solutions.

$$x\leq -3\text{; }x\geq \frac{1}{2}$$