Solve 3x^2-7x+8geq0

$$3 x ^ { 2 } - 7 x + 8 \geq 0$$

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.

$$3x^{2}-7x+8=0$$

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 3 for a, -7 for b, and 8 for c in the quadratic formula.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 3\times 8}}{2\times 3}$$

Do the calculations.

$$x=\frac{7±\sqrt{-47}}{6}$$

Since the square root of a negative number is not defined in the real field, there are no solutions. Expression 3x^{2}-7x+8 has the same sign for any x. To determine the sign, calculate the value of the expression for x=0.

$$3\times 0^{2}-7\times 0+8=8$$

The value of the expression 3x^{2}-7x+8 is always positive. Inequality holds for x\in \mathrm{R}.

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