Solve 0b20x^2-41x+20geq0

$$0b20 { x }^{ 2 } -41x+20 \geq 0$$

Solve for x

$$x\leq \frac{20}{41}$$

Solution Steps

Anything times zero gives zero.

$$0-41x+20\geq 0$$

Add 0 and 20 to get 20.

$$20-41x\geq 0$$

Subtract 20 from both sides. Anything subtracted from zero gives its negation.

$$-41x\geq -20$$

Divide both sides by -41. Since -41 is negative, the inequality direction is changed.

$$x\leq \frac{-20}{-41}$$

Fraction \frac{-20}{-41} can be simplified to \frac{20}{41} by removing the negative sign from both the numerator and the denominator.

$$x\leq \frac{20}{41}$$