Solve 3x^2+2x+4geq0

$$3 x ^ { 2 } + 2 x + 4 \geq 0$$

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.

$$3x^{2}+2x+4=0$$

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, 2 for b, and 4 for c in the quadratic formula.

$$x=\frac{-2±\sqrt{2^{2}-4\times 3\times 4}}{2\times 3}$$

Do the calculations.

$$x=\frac{-2±\sqrt{-44}}{6}$$

Since the square root of a negative number is not defined in the real field, there are no solutions. Expression 3x^{2}+2x+4 has the same sign for any x. To determine the sign, calculate the value of the expression for x=0.

$$3\times 0^{2}+2\times 0+4=4$$

The value of the expression 3x^{2}+2x+4 is always positive. Inequality holds for x\in \mathrm{R}.

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