Solve the Inequality for x 2x^2<=-x-4

$2 x^{2} \leq - x - 4$

Add $x$ to both sides of the inequality.

$2 x^{2} + x \leq - 4$

Add $4$ to both sides of the inequality.

$2 x^{2} + x + 4 \leq 0$

Convert the inequality to an equation.

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

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 2$ , $b = 1$ , and $c = 4$ into the quadratic formula and solve for $x$ .

$\frac{- 1 \pm \sqrt{1^{2} - 4 \cdot (2 \cdot 4)}}{2 \cdot 2}$

Simplify.

Simplify the numerator.

One to any power is one.

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 2 \cdot 4}}{2 \cdot 2}$

Multiply $- 4$ by $2$ .

$x = \frac{- 1 \pm \sqrt{1 - 8 \cdot 4}}{2 \cdot 2}$

Multiply $- 8$ by $4$ .

$x = \frac{- 1 \pm \sqrt{1 - 32}}{2 \cdot 2}$

Subtract $32$ from $1$ .

$x = \frac{- 1 \pm \sqrt{- 31}}{2 \cdot 2}$

Rewrite $- 31$ as $- 1 (31)$ .

$x = \frac{- 1 \pm \sqrt{- 1 \cdot 31}}{2 \cdot 2}$

Rewrite $\sqrt{- 1 (31)}$ as $\sqrt{- 1} \cdot \sqrt{31}$ .

$x = \frac{- 1 \pm \sqrt{- 1} \cdot \sqrt{31}}{2 \cdot 2}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{- 1 \pm i \sqrt{31}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{- 1 \pm i \sqrt{31}}{4}$

Identify the leading coefficient.

The leading term in a polynomial is the term with the highest degree.

$2 x^{2}$

The leading coefficient in a polynomial is the coefficient of the leading term.

$2$

Since there are no real x-intercepts and the leading coefficient is positive, the parabola opens up and $2 x^{2} + x$ is always greater than $0$ .

No solution