Solve the Inequality for x -5x^2-3x<=10x+6

$- 5 x^{2} - 3 x \leq 10 x + 6$

Move all terms containing $x$ to the left side of the inequality.

Subtract $10 x$ from both sides of the inequality.

$- 5 x^{2} - 3 x - 10 x \leq 6$

Subtract $10 x$ from $- 3 x$ .

$- 5 x^{2} - 13 x \leq 6$

Convert the inequality to an equation.

$- 5 x^{2} - 13 x = 6$

Subtract $6$ from both sides of the equation.

$- 5 x^{2} - 13 x - 6 = 0$

Factor the left side of the equation.

Factor $- 1$ out of $- 5 x^{2} - 13 x - 6$ .

Factor $- 1$ out of $- 5 x^{2}$ .

$- (5 x^{2}) - 13 x - 6 = 0$

Factor $- 1$ out of $- 13 x$ .

$- (5 x^{2}) - (13 x) - 6 = 0$

Rewrite $- 6$ as $- 1 (6)$ .

$- (5 x^{2}) - (13 x) - 1 \cdot 6 = 0$

Factor $- 1$ out of $- (5 x^{2}) - (13 x)$ .

$- (5 x^{2} + 13 x) - 1 \cdot 6 = 0$

Factor $- 1$ out of $- (5 x^{2} + 13 x) - 1 (6)$ .

$- (5 x^{2} + 13 x + 6) = 0$

Factor.

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 5 \cdot 6 = 30$ and whose sum is $b = 13$ .

Factor $13$ out of $13 x$ .

$- (5 x^{2} + 13 (x) + 6) = 0$

Rewrite $13$ as $3$ plus $10$

$- (5 x^{2} + (3 + 10) x + 6) = 0$

Apply the distributive property.

$- (5 x^{2} + 3 x + 10 x + 6) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$- ((5 x^{2} + 3 x) + 10 x + 6) = 0$

Factor out the greatest common factor (GCF) from each group.

$- (x (5 x + 3) + 2 (5 x + 3)) = 0$

Factor the polynomial by factoring out the greatest common factor, $5 x + 3$ .

$- ((5 x + 3) (x + 2)) = 0$

Remove unnecessary parentheses.

$- (5 x + 3) (x + 2) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$5 x + 3 = 0$

$x + 2 = 0$

Set $5 x + 3$ equal to $0$ and solve for $x$ .

Set $5 x + 3$ equal to $0$ .

$5 x + 3 = 0$

Solve $5 x + 3 = 0$ for $x$ .

Subtract $3$ from both sides of the equation.

$5 x = - 3$

Divide each term in $5 x = - 3$ by $5$ and simplify.

Divide each term in $5 x = - 3$ by $5$ .

$\frac{5 x}{5} = \frac{- 3}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 x}{5} = \frac{- 3}{5}$

Divide $x$ by $1$ .

$x = \frac{- 3}{5}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{3}{5}$

Set $x + 2$ equal to $0$ and solve for $x$ .

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

The final solution is all the values that make $- (5 x + 3) (x + 2) = 0$ true.

$x = - \frac{3}{5}, - 2$

Use each root to create test intervals.

$x < - 2$

$- 2 < x < - \frac{3}{5}$

$x > - \frac{3}{5}$

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

Test a value on the interval $x < - 2$ to see if it makes the inequality true.

Choose a value on the interval $x < - 2$ and see if this value makes the original inequality true.

$x = - 4$

Replace $x$ with $- 4$ in the original inequality.

$- 5 {(- 4)}^{2} - 3 \cdot - 4 \leq 10 (- 4) + 6$

The left side $- 68$ is less than the right side $- 34$ , which means that the given statement is always true.

True

Test a value on the interval $- 2 < x < - \frac{3}{5}$ to see if it makes the inequality true.

Choose a value on the interval $- 2 < x < - \frac{3}{5}$ and see if this value makes the original inequality true.

$x = - 1$

Replace $x$ with $- 1$ in the original inequality.

$- 5 {(- 1)}^{2} - 3 \cdot - 1 \leq 10 (- 1) + 6$

The left side $- 2$ is greater than the right side $- 4$ , which means that the given statement is false.

False

Test a value on the interval $x > - \frac{3}{5}$ to see if it makes the inequality true.

Choose a value on the interval $x > - \frac{3}{5}$ and see if this value makes the original inequality true.

$x = 0$

Replace $x$ with $0$ in the original inequality.

$- 5 {(0)}^{2} - 3 \cdot 0 \leq 10 (0) + 6$

The left side $0$ is less than the right side $6$ , which means that the given statement is always true.

True

Compare the intervals to determine which ones satisfy the original inequality.

$x < - 2$ True

$- 2 < x < - \frac{3}{5}$ False

$x > - \frac{3}{5}$ True

$x < - 2$ True

$- 2 < x < - \frac{3}{5}$ False

$x > - \frac{3}{5}$ True

The solution consists of all of the true intervals.

$x \leq - 2$ or $x \geq - \frac{3}{5}$

The result can be shown in multiple forms.

Inequality Form:

$x \leq - 2 or x \geq - \frac{3}{5}$

Interval Notation:

$(- \infty, - 2] \cup [- \frac{3}{5}, \infty)$

Solve the Inequality for x -5x^2-3x<=10x+6

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