Solve 8x^2-19x+20geq0

$$8 x ^ { 2 } - 19 x + 20 \geq 0$$

Solve for x

$$x\in \mathrm{R}$$

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.

$$8x^{2}-19x+20=0$$

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 8 for a, -19 for b, and 20 for c in the quadratic formula.

$$x=\frac{-\left(-19\right)±\sqrt{\left(-19\right)^{2}-4\times 8\times 20}}{2\times 8}$$

Do the calculations.

$$x=\frac{19±\sqrt{-279}}{16}$$

Since the square root of a negative number is not defined in the real field, there are no solutions. Expression 8x^{2}-19x+20 has the same sign for any x. To determine the sign, calculate the value of the expression for x=0.

$$8\times 0^{2}-19\times 0+20=20$$

The value of the expression 8x^{2}-19x+20 is always positive. Inequality holds for x\in \mathrm{R}.

$$x\in \mathrm{R}$$