Solve x^2/3+4/3x-7<0

$$\frac { x ^ { 2 } } { 3 } + \frac { 4 } { 3 } x - 7 < 0$$

Solve for x

$$x\in \left(-7,3\right)$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 3. Since 3 is positive, the inequality direction remains the same.

$$x^{2}+4x-21<0$$

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.

$$x^{2}+4x-21=0$$

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 -21 for c in the quadratic formula.

$$x=\frac{-4±\sqrt{4^{2}-4\times 1\left(-21\right)}}{2}$$

Do the calculations.

$$x=\frac{-4±10}{2}$$

Solve the equation x=\frac{-4±10}{2} when ± is plus and when ± is minus.

$$x=3 x=-7$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-3\right)\left(x+7\right)<0$$

For the product to be negative, x-3 and x+7 have to be of the opposite signs. Consider the case when x-3 is positive and x+7 is negative.

$$x-3>0 x+7<0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x+7 is positive and x-3 is negative.

$$x+7>0 x-3<0$$

The solution satisfying both inequalities is x\in \left(-7,3\right).

$$x\in \left(-7,3\right)$$

The final solution is the union of the obtained solutions.

$$x\in \left(-7,3\right)$$