Solve 2x^2-13x-34geq0

$$2 x ^ { 2 } - 13 x - 34 \geq 0$$

Solve for x

$$x\in (-\infty,-2]\cup [\frac{17}{2},\infty)$$

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.

$$2x^{2}-13x-34=0$$

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, -13 for b, and -34 for c in the quadratic formula.

$$x=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 2\left(-34\right)}}{2\times 2}$$

Do the calculations.

$$x=\frac{13±21}{4}$$

Solve the equation x=\frac{13±21}{4} when ± is plus and when ± is minus.

$$x=\frac{17}{2} x=-2$$

Rewrite the inequality by using the obtained solutions.

$$2\left(x-\frac{17}{2}\right)\left(x+2\right)\geq 0$$

For the product to be ≥0, x-\frac{17}{2} and x+2 have to be both ≤0 or both ≥0. Consider the case when x-\frac{17}{2} and x+2 are both ≤0.

$$x-\frac{17}{2}\leq 0 x+2\leq 0$$

The solution satisfying both inequalities is x\leq -2.

$$x\leq -2$$

Consider the case when x-\frac{17}{2} and x+2 are both ≥0.

$$x+2\geq 0 x-\frac{17}{2}\geq 0$$

The solution satisfying both inequalities is x\geq \frac{17}{2}.

$$x\geq \frac{17}{2}$$

The final solution is the union of the obtained solutions.

$$x\leq -2\text{; }x\geq \frac{17}{2}$$