Solve 772-2x^2+x-1leq0

$$772 - 2 x ^ { 2 } + x - 1 \leq 0$$

Solve for x

$$x\in (-\infty,\frac{1-\sqrt{6169}}{4}]\cup [\frac{\sqrt{6169}+1}{4},\infty)$$

Steps Using the Quadratic Formula

Subtract 1 from 772 to get 771.

$$771-2x^{2}+x\leq 0$$

Multiply the inequality by -1 to make the coefficient of the highest power in 771-2x^{2}+x positive. Since -1 is negative, the inequality direction is changed.

$$-771+2x^{2}-x\geq 0$$

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.

$$-771+2x^{2}-x=0$$

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, -1 for b, and -771 for c in the quadratic formula.

$$x=\frac{-\left(-1\right)±\sqrt{\left(-1\right)^{2}-4\times 2\left(-771\right)}}{2\times 2}$$

Do the calculations.

$$x=\frac{1±\sqrt{6169}}{4}$$

Solve the equation x=\frac{1±\sqrt{6169}}{4} when ± is plus and when ± is minus.

$$x=\frac{\sqrt{6169}+1}{4} x=\frac{1-\sqrt{6169}}{4}$$

Rewrite the inequality by using the obtained solutions.

$$2\left(x-\frac{\sqrt{6169}+1}{4}\right)\left(x-\frac{1-\sqrt{6169}}{4}\right)\geq 0$$

For the product to be ≥0, x-\frac{\sqrt{6169}+1}{4} and x-\frac{1-\sqrt{6169}}{4} have to be both ≤0 or both ≥0. Consider the case when x-\frac{\sqrt{6169}+1}{4} and x-\frac{1-\sqrt{6169}}{4} are both ≤0.

$$x-\frac{\sqrt{6169}+1}{4}\leq 0 x-\frac{1-\sqrt{6169}}{4}\leq 0$$

The solution satisfying both inequalities is x\leq \frac{1-\sqrt{6169}}{4}.

$$x\leq \frac{1-\sqrt{6169}}{4}$$

Consider the case when x-\frac{\sqrt{6169}+1}{4} and x-\frac{1-\sqrt{6169}}{4} are both ≥0.

$$x-\frac{1-\sqrt{6169}}{4}\geq 0 x-\frac{\sqrt{6169}+1}{4}\geq 0$$

The solution satisfying both inequalities is x\geq \frac{\sqrt{6169}+1}{4}.

$$x\geq \frac{\sqrt{6169}+1}{4}$$

The final solution is the union of the obtained solutions.

$$x\leq \frac{1-\sqrt{6169}}{4}\text{; }x\geq \frac{\sqrt{6169}+1}{4}$$