Solve x^2+5x+6>x

$$x ^ { 2 } + 5 x + 6 > x$$

Solve for x

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

Steps Using the Quadratic Formula

Subtract x from both sides.

$$x^{2}+5x+6-x>0$$

Combine 5x and -x to get 4x.

$$x^{2}+4x+6>0$$

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.

$$x^{2}+4x+6=0$$

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 1 for a, 4 for b, and 6 for c in the quadratic formula.

$$x=\frac{-4±\sqrt{4^{2}-4\times 1\times 6}}{2}$$

Do the calculations.

$$x=\frac{-4±\sqrt{-8}}{2}$$

Since the square root of a negative number is not defined in the real field, there are no solutions. Expression x^{2}+4x+6 has the same sign for any x. To determine the sign, calculate the value of the expression for x=0.

$$0^{2}+4\times 0+6=6$$

The value of the expression x^{2}+4x+6 is always positive. Inequality holds for x\in \mathrm{R}.

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