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

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

Solve for a

$$a\in \left(-\frac{1}{2},\frac{3}{2}\right)$$

Steps Using the Quadratic Formula

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

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

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.

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

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

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

Do the calculations.

$$a=\frac{4±8}{8}$$

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

$$a=\frac{3}{2} a=-\frac{1}{2}$$

Rewrite the inequality by using the obtained solutions.

$$4\left(a-\frac{3}{2}\right)\left(a+\frac{1}{2}\right)<0$$

For the product to be negative, a-\frac{3}{2} and a+\frac{1}{2} have to be of the opposite signs. Consider the case when a-\frac{3}{2} is positive and a+\frac{1}{2} is negative.

$$a-\frac{3}{2}>0 a+\frac{1}{2}<0$$

This is false for any a.

$$a\in \emptyset$$

Consider the case when a+\frac{1}{2} is positive and a-\frac{3}{2} is negative.

$$a+\frac{1}{2}>0 a-\frac{3}{2}<0$$

The solution satisfying both inequalities is a\in \left(-\frac{1}{2},\frac{3}{2}\right).

$$a\in \left(-\frac{1}{2},\frac{3}{2}\right)$$

The final solution is the union of the obtained solutions.

$$a\in \left(-\frac{1}{2},\frac{3}{2}\right)$$