Solve the Inequality for x -3x^2+2x+12>=7

$- 3 x^{2} + 2 x + 12 \geq 7$

Convert the inequality to an equation.

$- 3 x^{2} + 2 x + 12 = 7$

Subtract $7$ from both sides of the equation.

$- 3 x^{2} + 2 x + 12 - 7 = 0$

Subtract $7$ from $12$ .

$- 3 x^{2} + 2 x + 5 = 0$

Factor the left side of the equation.

Factor $- 1$ out of $- 3 x^{2} + 2 x + 5$ .

Factor $- 1$ out of $- 3 x^{2}$ .

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

Factor $- 1$ out of $2 x$ .

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

Rewrite $5$ as $- 1 (- 5)$ .

$- (3 x^{2}) - (- 2 x) - 1 \cdot - 5 = 0$

Factor $- 1$ out of $- (3 x^{2}) - (- 2 x)$ .

$- (3 x^{2} - 2 x) - 1 \cdot - 5 = 0$

Factor $- 1$ out of $- (3 x^{2} - 2 x) - 1 (- 5)$ .

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

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 $a \cdot c = 3 \cdot - 5 = - 15$ and whose sum is $b = - 2$ .

Factor $- 2$ out of $- 2 x$ .

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

Rewrite $- 2$ as $3$ plus $- 5$

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

Apply the distributive property.

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

$- (3 x (x + 1) - 5 (x + 1)) = 0$

Factor the polynomial by factoring out the greatest common factor, $x + 1$ .

$- ((x + 1) (3 x - 5)) = 0$

Remove unnecessary parentheses.

$- (x + 1) (3 x - 5) = 0$

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

$x + 1 = 0$

$3 x - 5 = 0$

Set $x + 1$ equal to $0$ and solve for $x$ .

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

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

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

$3 x - 5 = 0$

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

Add $5$ to both sides of the equation.

$3 x = 5$

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

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

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

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

The final solution is all the values that make $- (x + 1) (3 x - 5) = 0$ true.

$x = - 1, \frac{5}{3}$

Use each root to create test intervals.

$x < - 1$

$- 1 < x < \frac{5}{3}$

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

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 $x < - 1$ to see if it makes the inequality true.

Choose a value on the interval $x < - 1$ and see if this value makes the original inequality true.

$x = - 4$

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

$- 3 {(- 4)}^{2} + 2 (- 4) + 12 \geq 7$

The left side $- 44$ is less than the right side $7$ , which means that the given statement is false.

False

Test a value on the interval $- 1 < x < \frac{5}{3}$ to see if it makes the inequality true.

Choose a value on the interval $- 1 < x < \frac{5}{3}$ and see if this value makes the original inequality true.

$x = 0$

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

$- 3 {(0)}^{2} + 2 (0) + 12 \geq 7$

The left side $12$ is greater than the right side $7$ , which means that the given statement is always true.

True

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

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

$x = 4$

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

$- 3 {(4)}^{2} + 2 (4) + 12 \geq 7$

The left side $- 28$ is less than the right side $7$ , which means that the given statement is false.

False

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

$x < - 1$ False

$- 1 < x < \frac{5}{3}$ True

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

$x < - 1$ False

$- 1 < x < \frac{5}{3}$ True

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

The solution consists of all of the true intervals.

$- 1 \leq x \leq \frac{5}{3}$

The result can be shown in multiple forms.

Inequality Form:

$- 1 \leq x \leq \frac{5}{3}$

Interval Notation:

$[- 1, \frac{5}{3}]$

Solve the Inequality for x -3x^2+2x+12>=7

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