Solve (2a-1)^2-4(1-2a)>0

$$( 2 a - 1 ) ^ { 2 } - 4 ( 1 - 2 a ) > 0$$

Solve for a

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

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2a-1\right)^{2}.

$$4a^{2}-4a+1-4\left(1-2a\right)>0$$

Use the distributive property to multiply -4 by 1-2a.

$$4a^{2}-4a+1-4+8a>0$$

Subtract 4 from 1 to get -3.

$$4a^{2}-4a-3+8a>0$$

Combine -4a and 8a to get 4a.

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

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.

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

Do the calculations.

$$a=\frac{-4±8}{8}$$

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

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

Rewrite the inequality by using the obtained solutions.

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

For the product to be positive, a-\frac{1}{2} and a+\frac{3}{2} have to be both negative or both positive. Consider the case when a-\frac{1}{2} and a+\frac{3}{2} are both negative.

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

The solution satisfying both inequalities is a<-\frac{3}{2}.

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

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

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

The solution satisfying both inequalities is a>\frac{1}{2}.

$$a>\frac{1}{2}$$

The final solution is the union of the obtained solutions.

$$a<-\frac{3}{2}\text{; }a>\frac{1}{2}$$