Solve -3a^2-4a+4>0

$$- 3 a ^ { 2 } - 4 a + 4 > 0$$

Solve for a

$$a\in \left(-2,\frac{2}{3}\right)$$

Steps Using the Quadratic Formula

Multiply the inequality by -1 to make the coefficient of the highest power in -3a^{2}-4a+4 positive. Since -1 is negative, the inequality direction is changed.

$$3a^{2}+4a-4<0$$

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.

$$3a^{2}+4a-4=0$$

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 3 for a, 4 for b, and -4 for c in the quadratic formula.

$$a=\frac{-4±\sqrt{4^{2}-4\times 3\left(-4\right)}}{2\times 3}$$

Do the calculations.

$$a=\frac{-4±8}{6}$$

Solve the equation a=\frac{-4±8}{6} when ± is plus and when ± is minus.

$$a=\frac{2}{3} a=-2$$

Rewrite the inequality by using the obtained solutions.

$$3\left(a-\frac{2}{3}\right)\left(a+2\right)<0$$

For the product to be negative, a-\frac{2}{3} and a+2 have to be of the opposite signs. Consider the case when a-\frac{2}{3} is positive and a+2 is negative.

$$a-\frac{2}{3}>0 a+2<0$$

This is false for any a.

$$a\in \emptyset$$

Consider the case when a+2 is positive and a-\frac{2}{3} is negative.

$$a+2>0 a-\frac{2}{3}<0$$

The solution satisfying both inequalities is a\in \left(-2,\frac{2}{3}\right).

$$a\in \left(-2,\frac{2}{3}\right)$$

The final solution is the union of the obtained solutions.

$$a\in \left(-2,\frac{2}{3}\right)$$